Emitter resistor in BJT differential amplifier
I admit I'm not entirely sure of how well your words have communicated to me. And I'm very sure that I do not understand exactly what you need to hear from me. LvW has already provided a standard segue into it, although I've no idea how well that falls upon what you are struggling for. My only addition in that vein is to suggest that you carefully read this page. They cover your questions with all the necessary math and reasoning. If you work at that page, I think you can recover much of what you want.
But I often struggle for two things, simultaneously. One is how to develop quantitative models that predict well enough to be of practical use. The other is how to develop basic intuition -- a more qualitative approach that aids me somewhat when I skim over schematics, identifying the more important aspects of each functional grouping as well as I may, and then move on more quickly to search out what I'm looking for or stop at places that catch my interest and perhaps become the target for more thought.
So in that vein, let me start with what I think you are discussing in your question:

Gaining an idea about a first order approximation of the differential mode voltage gain given a single-ended source, with the other base tied to a voltage source that is at the same voltage as the reference point for the single-ended input source; and,
Gaining an idea about a first order approximation of the voltage gain for a common-mode change at both bases.

I'll start by adding a few details to your schematic:



simulate this circuit &ndash; Schematic created using CircuitLab
I've included the dynamic resistance (just the thermal voltage of the PN junction divided by the current through it) right at the tip of the BJT emitter. This is NOT part of your schematic. I'm just inserting it there to make it visible and hard to ignore. (Do keep in mind that this is only for AC analysis and not for DC operating point purposes, of course.)
Case 1 -- Applying 'emitter-degenerated CE amplifier' model
Here \$Q_2\$'s base (\$V_2\$) is tightly nailed down to a voltage source. So it cannot move. This also means that \$V_{\text{TIP}_2}\$ is similarly and equally nailed down, as well. (The idea of \$r_e\$ is to capture the effect of what happens at the emitter when a tiny change takes place across the base-emitter PN junction. So it is perfectly proper to consider \$V_{\text{TIP}_2}\$ as fixed and equally treatable as an ideal voltage.)
If \$R_\text{E}\$ were discounted (treated as open), then the analysis is pretty easy. You have \$R_\text{C}\$ in the collector and you have two \$r_e\$ values in the emitter leg. If you have done prior work on basic emitter-degenerated CE amplifier stages, then you know that the voltage gain magnitude is just the collector resistance divided by the emitter resistance. Here, and considering \$V_{\text{O}_1}\$ that would clearly mean, \$A_{v_d}\approx -\frac{R_\text{C}}{2\,r_e}\$.
Notice that the very important simplification made was to consider \$R_\text{E}\$ as an open (infinite resistance.)
Since we are talking only about the tiniest of changes, we can reasonably make the statement that \$V_\text{E}\$ doesn't change for these purposes. And if that doesn't change, then the current magnitude in \$R_\text{E}\$ also doesn't change. So we just assume it exactly doesn't change. But that assumption is the same as if \$R_\text{E}\$ were of infinite magnitude. (Think about this.)
This assumption isn't true. But it simplifies analysis. Putting its value back into analysis would result in a more complicated equation. But for many purposes, not a much better one.
The result here depends on dragging over an idea from the emitter-degenerated CE amplifier. But that's cheating because I just hauled over the equation and used it as if it was right to use.
But what if that's wrong? The following will avoid cheating.
Case 1 -- An analytic method
Let's use KCL and get the value for \$V_\text{E}\$:
$$\begin{align*}
\frac{V_\text{E}}{r_e}+\frac{V_\text{E}}{r_e} &amp;= I_{\text{R}_\text{E}}+\frac{V_{\text{TIP}_1}}{r_e}+\frac{V_{\text{TIP}_2}}{r_e}\\\\&amp;\therefore\\\\V_\text{E}&amp;=\frac{I_{\text{R}_\text{E}}\cdot r_e + V_{\text{TIP}_1}+V_{\text{TIP}_2}}{2}
\end{align*}$$
Now, we want to know how this value changes when \$V_1\$ changes. We can assume that the only part of the above equation that changes is \$V_{\text{TIP}_1}\$. So this means that: \$\Delta \,V_\text{E}\approx \frac12\,\Delta\,V_{\text{TIP}_1}\$.
(Keep in mind that \$V_{\text{TIP}_1}\$ is a fixed distance away from \$V_1\$. So this is the same thing as saying: \$\Delta \,V_\text{E}\approx \frac12\,\Delta\,V_{1}\$. Just FYI.)

We are allowed to shrink \$\Delta\,V_{1}\$ to an arbitrarily small
value. In fact, so small that while its magnitude is larger than zero
it is also smaller than the smallest finite value nearest to zero.

So, the change in \$Q_1\$'s emitter current will be:
$$\Delta \,I_{\text{E}_1} = \frac{\Delta\,V_{\text{TIP}_1}-\Delta \,V_\text{E}}{r_e}=\frac{\Delta\,V_{\text{TIP}_1}-\frac12\,\Delta\,V_{\text{TIP}_1}}{r_e}=\frac12\,\frac{\Delta\,V_{\text{TIP}_1}}{r_e}$$
Neglecting base currents, such that \$\Delta \,I_{\text{C}_1}=\Delta \,I_{\text{E}_1}\$, that tiny change then also causes a change in the voltage drop across \$Q_1\$'s \$R_\text{C}\$:
$$\Delta \,V_{\text{OUT}_1} = -\Delta \,I_{\text{C}_1}\cdot R_\text{C}=\frac12\,\frac{\Delta\,V_{\text{TIP}_1}}{r_e}\cdot R_\text{C}= \frac{R_\text{C}}{2\,r_e}\cdot \Delta\,V_{\text{TIP}_1}$$
Dividing both sides by \$\Delta\,V_{\text{TIP}_1}\$, find the differential mode gain as:
$$A_{v_d}=\frac{\Delta \,V_{\text{OUT}_1}}{\Delta\,V_{1}}=\frac{\Delta \,V_{\text{OUT}_1}}{\Delta\,V_{\text{TIP}_1}}=\frac{R_\text{C}}{2\,r_e}$$
That's the same result, but done analytically on the schematic.
Case 2 -- An analytic method
If I understand your wording (perhaps I'm projecting), then in this case you want to assume that \$V_1=V_2\$. (Both bases are set to the same voltage value to start.) Then observe the output voltages, \$V_{\text{O}_1}=V_{\text{O}_2}\$. (Both collectors should possess the same voltage value as a result.) Then apply a change to both inputs such that \$\Delta\,V_1=\Delta\,V_2\$ and work out how both \$V_{\text{O}_1}\$ and \$V_{\text{O}_2}\$ change in response.
In this case, we have one \$r_e\$ in parallel with another \$r_e\$, so for the voltage divider purposes we can merge these into a single effective dynamic resistance of \$\frac12\,r_e\$.
I'm going to skip the KCL solution process to get \$V_\text{E}\$ and instead move immediately to the change in \$V_\text{E}\$ for a given change, \$\Delta\,V_1=\Delta\,V_2=\Delta\,V_{\text{TIP}_1}=\Delta\,V_{\text{TIP}_2}\$.
The voltage divider computation can be used to work this out:
$$\Delta \,V_\text{E}=\frac{R_\text{E}}{R_\text{E}+\frac12\,r_e}\cdot \Delta\,V_{\text{TIP}_1}$$
That will incur a small change in \$I_{\text{R}_\text{E}}\$ and therefore: \$\Delta\,I_{\text{R}_\text{E}}=\frac{\Delta \,V_\text{E}}{R_\text{E}}\$. Again neglecting base currents so that \$\Delta \,I_{\text{C}_1}=\Delta \,I_{\text{E}_1}\$, that amount is split equally between the emitter currents for \$Q_1\$ and \$Q_2\$. It follows that \$\Delta \,I_{\text{C}_1}=\frac12\,\Delta\,I_{\text{R}_\text{E}}\$.
That tiny change then once again also causes a tiny change in the voltage drop across \$Q_1\$'s \$R_\text{C}\$:
$$\begin{align*}
\Delta\,V_{\text{O}_1}&amp;=-\Delta \,I_{\text{C}_1}\cdot R_\text{C}=-\frac12\,\Delta\,I_{\text{R}_\text{E}}\cdot R_\text{C}=-\frac12\,\frac{\Delta \,V_\text{E}}{R_\text{E}}\cdot R_\text{C}=-\frac12\,\frac{R_\text{C}}{R_\text{E}}\cdot \Delta \,V_\text{E}\\\\&amp;=-\frac12\,\frac{R_\text{C}}{R_\text{E}}\cdot\frac{R_\text{E}}{R_\text{E}+\frac12\,r_e}\cdot \Delta\,V_{\text{TIP}_1}\\\\&amp;=-\frac12\frac{R_\text{C}}{R_\text{E}+\frac12\,r_e}\cdot \Delta\,V_{\text{TIP}_1}
\end{align*}$$
Dividing both sides by \$\Delta\,V_{\text{TIP}_1}\$ and recognizing that \$\Delta\,V_{\text{TIP}_1}=\Delta\,V_{1}\$ find the common mode gain as:
$$A_{v_{cm}}=\frac{\Delta\,V_{\text{O}_1}}{\Delta\,V_{1}}=-\frac12\frac{R_\text{C}}{R_\text{E}+\frac12\,r_e}=-\frac{R_\text{C}}{2\,R_\text{E}+r_e}$$
If you examine your own answer and divide through by \$\beta\$ you'll find your answer remarkably similar to this one.
(The only real difference is the fact that I've neglected the base current. Feel free to re-add that back in to the above and work out the new result. But the difference is negligible and therefore beneath notice for these purposes.)
Summary
The above approaches are non-standard and not likely to be found in just the above way in some textbook. At the outset here I pointed you to a page that follows well-worn approaches for this problem and then I proceeded to avoid it.
I enjoy using these same standard approaches, myself. I don't mean to dissuade you from that path. I even provided a page to help, there. But I also want to show you that there are many other ways of seeing the same thing and that actively developing and continually refining this flexibility and creativity of mind is also important -- perhaps even more important. (I think it is much more important to be frequently insightful than it is to be frequently right about things. Insights can be tested and thrown away, when wrong. But they can take you to solutions no others may see, too.
So, as you learn the well-worn path, also force yourself to try and consider new ways to see the same problem. Develop seeing a problem from at least three different ways, if not more. Verify that all of them take you to a similar place. Learn which are good, throw the rest away. You will benefit from that effort. You may then be able to help others and point to solutions they can only see better once you have shown them how to see it.
It's worth some effort. I can say so from experience. Personal stories are important because they are true. And I'd tell you one or two, but it would be difficult to write them where the meaning I want --  to convey the advantages of seeing many paths -- comes across well and where I avoid unintended self-aggrandizing. I don't know how to write well enough to achieve that without a lot of effort, so I won't even start.