I'm interested in an asymptotic expansion of the following Riemann zeta-type function $$ \begin{align} \displaystyle \zeta(s \mid a,b) := \sum_{n=1}^{\infty} \frac{1}{(n+a)^{s}(n+b)}, \quad \Re a >-1, \, \Re b >-1, \, s>0, \tag1 \end{align} $$ as $s \to 0^+$.
The case $a=b$ in $(1)$ leads to the Riemann Hurwitz zeta function with the Laurent expansion near $0$:
$$
\begin{align} \sum_{n=1}^{\infty} \frac{1}{(n+a)^{s+1}}
= \frac{1}{s}+\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!} \gamma_{k}(a)s^{k}, \quad s>0, \tag2
\end{align}
$$$$
\begin{align} \sum_{n=1}^{\infty} \frac{1}{(n+a)^{s+1}}
= \frac{1}{s}+\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!} \gamma_{k}(a+1)s^{k}, \quad s>0, \tag2
\end{align}
$$ where $\displaystyle \gamma_{k}(a)$$\displaystyle \gamma_{k}(a+1)$ are the generalized Stieltjes constants with $\displaystyle \gamma_{0}(a)=-\Gamma'(a)/\Gamma(a)$$\displaystyle \gamma_{0}(a+1)=-\Gamma'(a+1)/\Gamma(a+1)$.
What is an asymptotic expansion, as $s \to 0^+$, of $\displaystyle \zeta(s \mid a,b)$ when $a\neq b$?