Deriving Differential Mode Gain of an Op Amp
You can certainly find it without using the superposition principle. 

If this is your opamp as a differential amplifier:







simulate this circuit &ndash; Schematic created using CircuitLab

You can always go use the following equation for the output voltage:

$$V_o=A(V^+-V^-)$$

Where \$A\$ is the open loop gain (for an ideal opamp \$A\rightarrow\infty\$).

Now you need to find \$V^-\$ and \$V^+\$ in order to plug them in the previous equation and find an equation for the output voltage. 

Recall that for an ideal opamp, the input current is zero, therefore for \$V^+\$:

$$ V^+=\frac{R_4}{R_3+R_4}V_1$$

For \$V^-\$, 

$$ V^-=\frac{R_2}{R_1+R_2}V_2+\frac{R_1}{R_1+R_2}V_o$$

You can now plug those into the \$V_o=A(V^+-V^-)\$ equation:

$$V_o=A\bigg(\frac{R_4}{R_3+R_4}V_1-\frac{R_2}{R_1+R_2}V_2-\frac{R_1}{R_1+R_2}V_o\bigg) $$

$$V_o\bigg(1+A\frac{R_1}{R_1+R_2}\bigg)=A\bigg(\frac{R_4}{R_3+R_4}V_1-\frac{R_2}{R_1+R_2}V_2\bigg) $$

$$V_o\bigg(\frac{R_1+R_2+AR_1}{R_1+R_2}\bigg)=A\bigg(\frac{R_4}{R_3+R_4}V_1-\frac{R_2}{R_1+R_2}V_2\bigg) $$

$$V_o=A\bigg(\frac{R_1+R_2}{R_1+R_2+AR_1}\bigg)\bigg(\frac{R_4}{R_3+R_4}V_1-\frac{R_2}{R_1+R_2}V_2\bigg) $$

$$V_o=\frac{A}{A}\bigg(\frac{R_1+R_2}{\frac{R_1+R_2}{A}+R_1}\bigg)\bigg(\frac{R_4}{R_3+R_4}V_1-\frac{R_2}{R_1+R_2}V_2\bigg) $$

$$V_o=\bigg(\frac{R_1+R_2}{\frac{R_1+R_2}{A}+R_1}\bigg)\bigg(\frac{R_4}{R_3+R_4}V_1-\frac{R_2}{R_1+R_2}V_2\bigg) $$

Recall that for an ideal opamp \$A\rightarrow\infty\$, then \$\frac{R_1+R_2}{A}\rightarrow0\$

$$V_o\approx\bigg(\frac{R_1+R_2}{R_1}\bigg)\bigg(\frac{R_4}{R_3+R_4}V_1-\frac{R_2}{R_1+R_2}V_2\bigg) $$

And finally, 

$$V_o\approx\bigg(\frac{R_1+R_2}{R_1}\bigg)\bigg(\frac{R_4}{R_3+R_4}\bigg)V_1-\bigg(\frac{R_2}{R_1}\bigg)V_2 $$

From there you can express it in terms of the differential input, \$V_d\$, and the common mode, \$V_{cm}\$, just as they do in the link you provided.

EDIT: Finding \$V^-\$

There are several ways you could find this. Let's try KCL:


Consider the node at \$V^-\$, there are two currents related to it. 
Let's assume the direction of those currents is into the node \$V^-\$
Then, there is one current that flows out of \$V_2\$, through \$R_1\$ into \$V^-\$ (let's call this \$I_1\$) and another that flows out \$V_o\$, through \$R_2\$ into \$V^-\$ (let's call this \$I_2\$).


Now, you can say, for the node at \$V^-\$:

$$ I_1+I_2=0$$
$$ \frac{V_2-V^-}{R_1}+\frac{V_o-V^-}{R_2}=0$$

That is just KCL. You can solve this for \$V^-\$:
$$ \frac{V_2}{R_1}-\frac{V^-}{R_1}+\frac{V_o}{R_2}-\frac{V^-}{R_2}=0$$

$$ \frac{V_2}{R_1}+\frac{V_o}{R_2}=V^-\bigg(\frac{1}{R_1}+\frac{1}{R_2}\bigg)$$
$$ \frac{V_2}{R_1}+\frac{V_o}{R_2}=V^-\bigg(\frac{R_1+R_2}{R_1R_2}\bigg)$$

Which leads to:

$$ V^-=\frac{R_2}{R_1+R_2}V_2+\frac{R_1}{R_1+R_2}V_o$$