It is known that: $$ (\nabla^2+k^2)(-\frac{e^{ikr}}{4\pi r})=\delta(\vec{r}) $$ where $k>0$ and $\delta(\vec{r})$ is the three dimensional Dirac delta function.
My question is, is it possible to find a function $f(\vec{r})$ that satisfies the following relation: $$ \left[\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}-(\frac{\partial}{\partial z}+i\alpha)^2-k^2\right] f(\vec{r})=\delta(\vec{r}) $$
Where $\alpha>0,k>0$.
Fourier transforming your equation, you can check that the function $f({\bf r})$ is given by $$f({\bf r}) = \int \frac{d^3 q}{(2\pi)^3} \frac{e^{i {\bf q} \cdot {\bf r}}}{(q_3+\alpha)^2 - q_1^2 -q_2^2 -k^2 }$$ with ${\bf q} = (q_1,q_2,q_3)$. You can still impose boundary conditions on $f({\bf r})$ which corresponds to shifting the position of the poles below or above the real axis (here, I will choose outgoing wave Green's function).
The integrand has poles at $q_3 = -\alpha \pm \sqrt{k^2 +q_1^2+q_2^2}$. We perform the integral over $q_3$ by closing the contour on the upper complex plane and taking the pole with $+$ inside the contour. We obtain $$f({\bf r}) = -\frac{i e^{i \alpha r_3}}{2} \int\frac{d^2 q}{(2\pi)^2} \frac{e^{i q_1 r_1 + i q_2 r_2 +i (k^2 + q_1^2 + q_2^2)^{1/2}r_3}}{\sqrt{k^2+q_1^2 + q_2^2}} .$$
For the remaining integral, we introduce $\rho = \sqrt{r_1^2+r_2^2}$ and $q_1 = q \cos \phi, q_2= q \sin\phi$, where $\phi$ is the angle measured with respect to $(r_1,r_2)$. We obtain $$\int\frac{d^2 q}{(2\pi)^2} \frac{e^{i q_1 r_1 + i q_2 r_2+i (k^2 + q_1^2 + q_2^2)^{1/2}r_3}}{\sqrt{k^2+q_1^2 + q_2^2}} = \int_0^\infty \!\frac{dq}{2\pi}\,\int_{0}^{2\pi}\!\frac{d\phi}{2\pi}\frac{q e^{iq \rho \cos\phi + i (k^2+q^2)^{1/2} r_3}}{\sqrt{k^2+q^2}}\\=\int_0^\infty \!\frac{dq}{2\pi} \frac{q e^{i (k^2+q^2)^{1/2} r_3} J_0(\rho q)}{\sqrt{k^2+q^2}} = -\frac{i e^{ik (r_3^2 -\rho^2)^{1/2}} }{2\pi \sqrt{r_3^2 -\rho^2}}. $$
So, we obtain the final result $$ f({\bf r}) =- \frac{ e^{-i \alpha r_3 +ik (r_3^2 -r_1^2-r_2^2)^{1/2}}}{4 \pi\sqrt{r_3^2 -r_1^2-r_2^2}}.$$
Edit:
There is an alternative much more conceptual way to arrive at the final result. It is obtained by using ideas of analytical continuation. In physics it is also known as Wick rotation. We are interested in a solution to $$\left[\partial_1^2+\partial_2^2-(\partial_3+i\alpha)^2-k^2\right] f({\bf r})=\delta({\bf r}).\tag{1}$$ For convenience, we first introduce a new function $\tilde f({\bf r}) = e^{-i \alpha r_3} f({\bf r})$ in terms of which (1) assumes the (homogeneous) form $$ (\partial_1^2+ \partial_2^2 - \partial_3^2 -k^2) \tilde f({\bf r}) =\delta({\bf r}).$$
Let us instead study the family of equations $$(-e^{i\theta}\partial_1^2-e^{i\theta} \partial_2^2 - \partial_3^2 -k^2) g_\theta({\bf r}) = \delta({\bf r}).$$ For $\theta=0$ we have $g_0({\bf r}) =e^{i k |{\bf r}|}/(4\pi |{\bf r}|)$ according to the question. We obtain $\tilde f$ from $g_\theta$ by analytical continuation as $\theta \uparrow \pi$ or $\theta \downarrow -\pi$.
We have the general relation $$ g_\theta({\bf r}) = \int\!\frac{d^3q}{(2\pi)^3} \frac{e^{i{\bf q}\cdot{\bf r}}}{q_3^3 + e^{i\theta} (q_1^2 + q_2^2) -k^2}. $$ Introducing new integration variable $ \tilde{\bf q}= (e^{i\theta/2}q_1,e^{i\theta/2}q_2,q_3)$, we obtain the alternative expression (here we have to be careful that there is no contribution from the arc at infinity) $$ g_\theta({\bf r}) = e^{-i\theta}\int\!\frac{d^3 \tilde q}{(2\pi)^3} \frac{e^{i\tilde{\bf q}\cdot\tilde{\bf r}}}{\tilde {\bf q}^2 -k^2} =e^{-i\theta}g_0(\tilde{\bf r}) $$ with $$\tilde{\bf r}= (e^{-i\theta/2} r_1,e^{-i\theta/2} r_2 ,r_3).$$ Letting $\theta \to \pi$, we obtain the result $$ \tilde f({\bf r}) = g_\pi({\bf r}) = -g_0(-i r_1, -i r_2,r_3)= - \frac{e^{i k (r_3^2-r_1^2-r_2^2)^{1/2}}}{4\pi \sqrt{r_3^2-r_1^2-r_2^2}}$$
