= Solution
To implement <crossing symmetry> with the annihilation labels, replace the old scattering <four-momenta> by $(p_1,-p_3,-p_2,p_4)$. The scattering <Mandelstam variables> then become $(s_{\rm old},t_{\rm old},u_{\rm old})=(t,s,u)$, where now $s=(p_1+p_2)^2$, $t=(p_1-p_3)^2$ and $u=(p_1-p_4)^2$. Consequently
$$
\overline{|\mathcal M_{\rm ann}|^2}=2e^4\frac{t^2+u^2}{s^2}.
$$
In the <centre-of-momentum frame>, take $\theta$ between the incoming <electron> and outgoing positive <muon>, $p_3$. For negligible <masses>, $t=-s(1-\cos\theta)/2$ and $u=-s(1+\cos\theta)/2$. Therefore
$$
\boxed{\overline{|\mathcal M_{\rm ann}|^2}=e^4(1+\cos^2\theta).}
$$
Defining $\theta$ using the outgoing negative <muon> interchanges $t$ and $u$ and leaves the result unchanged.
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