 Solution

First fix the source at the spatial origin. Fourier transform replaces ${\partial }_t$ by $-i\omega$ and $\nabla$ by $i\mathbf{k}$ in the free term. Multiplication by a static background becomes a momentum convolution; the derivative in $A\cdot \partial G$ acts on the Green function, so it carries the internal momentum ${\mathbf{k}}^{\prime }$. ConsequentlyThe transformed point source supplies the right-hand side one. The signs follow from $A^{\mu }{\partial }_{\mu }=A^0{\partial }_t+\mathbf{A}\cdot \nabla$ in these conventions.

Source-position qualification. For a nonuniform $A(\mathbf{r})$, the general Green function is not translation invariant. In the printed basis $e^{i\mathbf{k}\cdot (\mathbf{r}-{\mathbf{r}}^{\prime })}$, its coefficient must also depend on ${\mathbf{r}}^{\prime }$, and the convolution contains an extra factor $e^{i(\mathbf{k}-{\mathbf{k}}^{\prime })\cdot {\mathbf{r}}^{\prime }}$. The displayed source equation is valid for ${\mathbf{r}}^{\prime }=0$, or with the background itself expressed in coordinates centred on that source. Equivalently, a general spatial Fourier Green function has two momentum arguments and a momentum-delta-function source. Staticity alone permits the time-difference transform, not a general position-difference-only transform.