I'm thinking this has to do with dominated convergence.
I know it kinda works, but I'm not 100% sure why and how the steps are rigorously carried out and possibly the above theorem is applied. Let me start:
Starting with the expression $$ f(x)=\sum_{k=1}^{\infty} k^{-s} \left(k+x\right)^{-s} $$ where $x>0$ and we see by inspection, that it clearly converges for $s>1/2$.
Now I'm interested in the large $x$ behaviour, so thinking inside the sum $k$ is fixed and $x$ large enough such that $\frac{k}{x}<1$ we expand $\left(k+x\right)^{-s}$ to obtain $$ =x^{-s} \sum_{k=1}^{\infty} k^{-s} \sum_{m=0}^\infty \begin{pmatrix} -s \\ m \end{pmatrix} \left(\frac{k}{x}\right)^m \\ =x^{-s} \sum_{k=1}^{\infty} \sum_{m=0}^\infty \begin{pmatrix} -s \\ m \end{pmatrix} x^{-m} \, k^{-\left(s-m\right)} \, . $$ Now I argue that for $k/x<c<1$ the inner sum is uniformly convergent and I can interchange summation order. What I'm not really clear about is, that the outer sum $k$ obviously goes to $\infty$, but for the matter of the inner sum I can view $k$ being fixed for the moment, so that summation order can be swapped?
Doing so we arrive at $$ =x^{-s} \sum_{m=0}^{\infty} \begin{pmatrix} -s \\ m \end{pmatrix} x^{-m} \sum_{k=1}^\infty k^{-\left(s-m\right)} \, . $$ But now in the same spirit $m$ is being viewed as fixed value and the inner sum converges for $s>m+1$. The final result is thus $$ f(x)=x^{-s} \sum_{m=0}^{\infty} \begin{pmatrix} -s \\ m \end{pmatrix} \zeta\left(s-m\right) \, x^{-m} $$ which is only an asymptotic series, because $\zeta(s-m)$ grows like $\frac{\Gamma(1+m-s)}{(2\pi)^{1+m-s}}$.
This is most likely due to the summation order interchangement. But does this also mean, that this function does not have a proper convergent asymptotic expansion?