In a Wilson mirror, does Q3 need to have the same beta to cancel the base current error?
I'm not sure how accurate you want to go, but there are a few things in this circuit that I believe will not make the current in both branches equal.

While \$V_{BE1} = V_{BE2}\$ means that \$I_{B1} = I_{B2}\$, in general \$V_{CE1} \neq V_{CE2}\$. This will generate a small difference in \$I_C\$ due to the Early effect.

Ignoring that difference, we can still find

$$I_p = I_{C1} + I_{B3} = \beta_1I_{B1} + I_{B3} \\
I_L = I_{C3} = \beta_3I_{B3}$$

We can also find an error in your 4th point, the KCL needs to be:

$$\begin{align}
I_{E3} &amp;= I_{C2} + I_{B1} + I_{B2} \\
&amp;\Downarrow \\
I_{C3} + I_{B3} &amp;= I_{C2} + 2I_{B1,2} \\
&amp;\Downarrow \\
I_{B3} &amp;= \frac{\beta_2 + 2}{\beta_3 + 1}I_{B1,2}
\end{align}$$

This means that point 2 and 3 are incorrect as well.

So finally

$$I_p = \left(\beta_1 + \frac{\beta_2 + 2}{\beta_3 + 1}\right)I_{B1,2} \\
I_L = \beta_3\frac{\beta_2 + 2}{\beta_3 + 1}I_{B1,2}$$

These two current are equal only if

$$\beta_1(\beta_3+1) + (\beta_2 + 2) = \beta_3(\beta_2 + 2)$$

Assuming \$\beta_{1,2} = \beta_1 = \beta_2\$ we can solve that

$$\beta_3 = \beta_{1,2} + 1$$

which turns out to be the same equation anyway. Now you can continue by observing that \$\beta \gg 1\$, such that \$\beta_3 \approx \beta_{1,2}\$.