= Solution
Let $p_1,p_2$ be the incoming scalar and antiscalar momenta and $k_1,k_2$ the outgoing photon momenta, so $p_1+p_2=k_1+k_2$. Up to the common overall phase fixed by the $S$-matrix convention, the connected leading amplitude is
$$
\mathcal M=e^2\epsilon_{1\mu}^*\epsilon_{2\nu}^*\mathcal M^{\mu\nu},
$$
where
$$
\mathcal M^{\mu\nu}
=2\eta^{\mu\nu}
+\frac{(2p_1-k_1)^\mu(2p_2-k_2)^\nu}{(p_1-k_1)^2-m^2}
+\frac{(2p_2-k_1)^\mu(2p_1-k_2)^\nu}{(p_1-k_2)^2-m^2}.
$$
On shell, the denominators are $-2p_1\cdot k_1=-2p_2\cdot k_2$ and $-2p_1\cdot k_2=-2p_2\cdot k_1$. Contracting with $k_{1\mu}$ therefore gives
$$
k_{1\mu}\mathcal M^{\mu\nu}epsilon_{2\nu}^*
=2k_1\cdot\epsilon_2^*
-(2p_2-k_2)\cdot\epsilon_2^*
-(2p_1-k_2)\cdot\epsilon_2^*=0,
$$
using momentum conservation and $k_2\cdot\epsilon_2=0$. The same calculation with the photons interchanged gives $k_{2\nu}\mathcal M^{\mu\nu}\epsilon_{1\mu}^*=0$. The two exchange diagrams and the seagull term are all required for this <Ward identity>.
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