the following paper says if $f(\vec{x})=f(x)$ does only depend on $x$ then we have
$\int d^3x' \frac{e^{\pm ik_{\phi}|\vec{x}-\vec{x}'|}}{|\vec{x}-\vec{x}'|} f(\vec{x}')= \frac{2\pi i}{k_{\phi}} e^{ ik_{\phi}x} \int dx' e^{-ik_{\phi}x'}f(x')$
You can find this by going from (3) to (4) in the linked paper. Further reduced this means that one can integrate out the two dimensional integral and obtain a one dimensional integral. In other words:
$\int dy'dz' \frac{e^{\pm ik_{\phi}|\vec{x}-\vec{x}'|}}{|\vec{x}-\vec{x}'|}=\frac{2\pi i}{k_{\phi}}e^{ik_{\phi}(x-x')} ~~~ (I)$
I tried to recalculate equation $(I)$ but got stuck at one point:
I substitute $z-z'=\alpha$, $(x-x')^2+(y-y')^2=\zeta^2$ and finally $\frac{\alpha}{\zeta}=\sinh\beta$. We get then $|\vec{x}-\vec{x}'|=\zeta\cosh\beta$. The double integral in $(I)$ gets:
$\int dy'dz' \frac{e^{\pm ik_{\phi}|\vec{x}-\vec{x}'|}}{|\vec{x}-\vec{x}'|}=\int dy' \zeta\int d\beta e^{\pm ik_{\phi}\zeta\cosh\beta}$
For the last integral I had the idea to shift the integration along the imaginary axis, since there are no poles. If I do that I get
$-i\int dy' \zeta\int_{-\infty}^{\infty} d\kappa e^{\pm ik_{\phi}\zeta\cos\kappa}$
This is almost a Bessel function, but the integration range does not fit. I still have to integrate from $-\infty$ to $\infty$, while I need a $2\pi$ integration range to identify a Bessel function.
I also tried out mathematica, but it did not give me an answer.
Many thanks in advance!