= Solution
The delta function in the artificial constant potential collapses the convolution. Put $b=e\gamma$ and $K^2=k^2+m^2$. Then
$$
\boxed{G(\mathbf k;\omega)=\frac1{-\omega^2+2ib\omega+K^2},
\qquad\omega_\pm=ib\pm\sqrt{K^2-b^2}.}
$$
Inverse <Fourier transform> gives the stated real-frequency integral with exponential $e^{-i\omega(t-t')}$. For $K^2>b^2$, the <poles> have opposite real parts and common imaginary part $b$. For $0<K^2<b^2$, both are imaginary, at $i[b\pm\sqrt{b^2-K^2}]$. They remain in the upper half-plane if $b>0$, or the lower half-plane if $b<0$. At equality they merge into a \b[second-order <pole> at $ib$]. For $K^2=0$ one <pole> reaches the real axis, requiring a prescription.
\Image[/past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2014/paper-1-tp1-poles.png]
{title=Pole motion above, at and below the repeated-pole threshold; shown for positive e gamma}
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