Following @Abdelmalek's great advice in the comments above it is enough to compute the following triple integral:
\begin{align*}
I=\frac{\pi^s}{\Gamma(s)}\int_{0}^{\infty}\left(\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}   
(x_1+x_2 i)^{\ell}\cdot e^{-\pi t(x_1^2+x_2^2+a^2)} e^{2\pi i (x_1\zeta_1+x_2\zeta_2)} dx_1 dx_2 \right)                                                      
t^s\frac{dt}{t}.
\end{align*}
So we have
\begin{align*}
I=&=\frac{\pi^s}{\Gamma(s)}\int_{0}^{\infty}t^{-1}\cdot\left(\frac{i}{t}(\zeta_1+i\zeta_2)\right)^{\ell}\cdot 
e^{-\frac{\pi}{t}(\zeta_1^2+\zeta_2^2)}\cdot e^{-\pi ta^2}t^{s}\frac{dt}{t}\\
&=\frac{(i)^{\ell}\pi^s}{\Gamma(s)}(\zeta_1+i\zeta_2)^{\ell}\int_{0}^{\infty}
e^{-\frac{\pi}{t}(\zeta_1^2+\zeta_2^2)-\pi ta^2}\cdot t^{s-\ell-1} \frac{dt}{t} \\
&=\frac{(i)^{\ell}\pi^s}{\Gamma(s)}(\zeta_1+i\zeta_2)^{\ell}\cdot 2\left(\frac{|\zeta|}{a}\right)^{s-\ell-1}
K_{s-\ell-1}\left(2\pi|\zeta|a\right)
\end{align*}
where $\zeta=\zeta_1+i\zeta_2$.

So modulo some minor mistakes made above, out of excitement, it means that the initial messy formulas that I had obtained could be vastly simplified, as what I was hoping for. In any case this new approach is just better and simpler!

I can't wait for somebody to publish a book with tables of double integrals (or even multiple integrals)!