Let's have minimally extended gauge invariant lagrangian (with free kinetic term of EM field): $$ \tag 1 L (\Psi , \partial_{\mu} \Psi) \to L (\Psi , D_{\mu}\Psi ) - \frac{1}{4}F^{\mu \nu}F_{\mu \nu}, \quad D_{\mu} = \partial_{\mu} - iQA_{\mu}. $$ Here $A_{\mu}$ is given by $F_{\mu \nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}$ .
It is well-known result that for an arbitrary amplitude of EM process (I have cut here photon propagators and polarization vectors; all of operators here is given in Heisenberg picture) $$ \tag 2 M^{\mu_{1}...\mu_{n}}(q_{1}, ..., q_{n}) = \int d^{4}x_{1}...d^{4}x_{n}e^{iq_{1}x_{1} + ...+iq_{n}x_{n}} \langle | \hat{T}\left( \hat{J}^{\mu_{1}}(x_{1})...\hat{J}^{\mu_{n}}(x_{n}) \right)|\rangle , $$ $$ \hat{J}^{\mu} = \hat{\bar{\Psi}}\gamma^{\mu}\hat{\Psi}, \quad \partial_{\mu}\hat{J}^{\mu} = 0, $$ we have identity $$ \tag 3 q_{\mu_{i}}M^{...\mu_{i}...} = 0. $$ Let's have than non-zero photon mass. Then we will get modified photon propagators (so the new force law) and non-invariance of lagrangian, but $\hat{J}_{\mu}$ is still conserved, so it seems that $(2)$ still satisfies identity $(3)$. I have checked this statement for processes $WW^{+} \to \gamma \gamma$ and $e e^{+} \to \gamma \gamma$ at the lowest order.
But is this statement true in general? If yes, why we usually say that photon must have strictly zero mass (nonzero but very small mass doesn't contradict the experimental data)? I'm not interested in Higgs mechanism, because I'm not interested in exactly gauge invariance.
