= Solution
Time independence of $V$ makes the <Green function> depend on $t,t'$ through $\tau=t-t'$. A spatially varying potential does not imply spatial <translation> invariance. Fix $\mathbf r'$ and set $\mathbf x=\mathbf r-\mathbf r'$; write the Fourier coefficient more fully as $G_{\mathbf r'}(\mathbf k;\omega)$. The <source> position is an implicit parameter in the printed notation. Define
$$
V_{\mathbf r'}(\mathbf q)=\int d^3x\,V(\mathbf r'+\mathbf x)e^{-i\mathbf q\cdot\mathbf x}=e^{i\mathbf q\cdot\mathbf r'}\widetilde V(\mathbf q).
$$
The ordinary derivative terms transform into $(-\omega^2+k^2+m^2)G_{\mathbf r'}$. The product $V\partial_tG$ becomes a <convolution>, and $2ie(-i\omega)=2e\omega$. More explicitly, insert both inverse spatial <Fourier transforms> into $VG$: the integral over $\mathbf x$ enforces $\mathbf q+\mathbf k'=\mathbf k$. The transformed delta <source> is one, leaving
$$
\boxed{(-\omega^2+k^2+m^2)G_{\mathbf r'}(\mathbf k;\omega)+2e\omega\int\frac{d^3k'}{(2\pi)^3}V_{\mathbf r'}(\mathbf k-\mathbf k')G_{\mathbf r'}(\mathbf k';\omega)=1.}
$$
This is a <source-position-dependent Fourier Green function>. Dropping the explicit $\mathbf r'$ subscripts recovers the requested notation with exactly the potential transform defined in the source. No assumption of a homogeneous potential is needed. The factor $(2\pi)^{-3}$ and the source-position phase are necessary for the stated Fourier convention.
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