How to meromorphically continue $$f(s)=\sum^\infty_{n=1}\sum^\infty_{m=1}\frac1{(an+bm)^s}$$ to $\Re s\le 2$? ($a,b$ are positive constants.)
In the case $a=b$, we have $$f(s)=a^{-s} \sum^\infty_{n=1}\sum^\infty_{m=1}\frac1{(n+m)^s} =a^{-s}\sum^\infty_{N=1}\frac{N-1}{N^s}=a^{-s}(\zeta(s-1)-\zeta(s))$$ and we can exploit the well-known analytic continuation of $\zeta$ to continue $f$ to a meromorphic function on $\mathbb C$.
How about for arbitrary $a,b$?
As usual, for $a >0,b>0,\Re(s) > 2$ $$\Gamma(s)F(a,b,s)=\Gamma(s)\sum_{n,m} (an+bm)^{-s}=\sum_{n,m} \int_0^\infty x^{s-1}e^{-(an+bm)x}dx$$ $$= \int_0^\infty x^{s-1}\sum_{n,m}e^{-(an+bm)x}dx=\int_0^\infty \frac{x^{s-1}}{(e^{ax}-1)(e^{bx}-1)}dx $$ The latter converges and is analytic for $\Re(a)> 0,\Re(b) > 0,\Re(s) > 2$.
With $f_{a,b}(x)=\frac{x^2}{(e^{ax}-1)(e^{bx}-1)}$ we obtain for $\Re(s) > 1-K$ $$\int_0^\infty x^{s-3}(f_{a,b}(x)-1_{x <1}\sum_{k\le K}\frac{f_{a,b}^{(k)}(0)}{k!}x^k)dx = \Gamma(s)F(a,b,s)-\sum_{k\le K}\frac{f_{a,b}^{(k)}(0)}{k!}\frac1{s-2+k}$$ which is the analytic continuation of the RHS to $\Re(s) > 1-K$.
Thus $\Gamma(s)F(a,b,s)$ is meromorphic on the whole complex plane and the integers $\le 2$ are simple poles of residue $\frac{f_{a,b}^{(2-k)}(0)}{(2-k)!}$,