For $s > 0$ we have

$$ \sum_{n=1}^{\infty} \frac{1}{(n+a)^s(n+b)} - \sum_{n=1}^{\infty} \frac{1}{(n+a)^{s+1}} = \sum_{n=1}^{\infty} \frac{a-b}{(n+a)^{s+1}(n+b)} =: g(s). $$

The series on the right-hand side converges and is analytic on $\operatorname{Re} s > -1$, so the difference on the left-hand side can be analytically continued to this region. Consequently, the analytic continuation $\zeta(s \mid a,b)$ of your sum satisfies

$$ \zeta(s \mid a,b) = \zeta(s+1,a+1) + g(s) $$

for $\operatorname{Re} s > -1$, $s \neq 0$ and thus has a simple pole at $s=0$ with residue $1$.

For $|s| < 1$ we have

$$ g(s) = \sum_{k=0}^\infty \left( \sum_{n=1}^\infty \frac{[\log(n+a)]^k}{(n+a)(n+b)} \right) \frac{(-s)^k}{k!}, $$

so for $0 < |s| < 1$

$$ \zeta(s \mid a,b) = \frac{1}{s} + \sum_{k=0}^{\infty} \left( \gamma_{k}(a+1) + \sum_{n=1}^\infty \frac{[\log(n+a)]^k}{(n+a)(n+b)} \right) \frac{(-s)^k}{k!}. $$

For $s > 0$ we have

$$ \sum_{n=1}^{\infty} \frac{1}{(n+a)^s(n+b)} - \sum_{n=1}^{\infty} \frac{1}{(n+a)^{s+1}} = \sum_{n=1}^{\infty} \frac{a-b}{(n+a)^{s+1}(n+b)} =: g(s). $$

The series on the right-hand side converges and is analytic on $\operatorname{Re} s > -1$, so the difference on the left-hand side can be analytically continued to this region. Consequently, the analytic continuation $\zeta(s \mid a,b)$ of your sum satisfies

$$ \zeta(s \mid a,b) = \zeta(s+1,a+1) + g(s) $$

for $\operatorname{Re} s > -1$, $s \neq 0$ and thus has a simple pole at $s=0$ with residue $1$.

For $|s| < 1$ we have

$$ g(s) = (a-b)\sum_{k=0}^\infty \left( \sum_{n=1}^\infty \frac{[\log(n+a)]^k}{(n+a)(n+b)} \right) \frac{(-s)^k}{k!}, $$

so for $0 < |s| < 1$

$$ \zeta(s \mid a,b) = \frac{1}{s} + \sum_{k=0}^{\infty} \left( \gamma_{k}(a+1) + (a-b)\sum_{n=1}^\infty \frac{[\log(n+a)]^k}{(n+a)(n+b)} \right) \frac{(-s)^k}{k!}. $$

For $s > 0$ we have

$$ \sum_{n=1}^{\infty} \frac{1}{(n+a)^s(n+b)} - \sum_{n=1}^{\infty} \frac{1}{(n+a)^{s+1}} = \sum_{n=1}^{\infty} \frac{a-b}{(n+a)^{s+1}(n+b)} =: g(s). $$

The series on the right-hand side converges and is analytic on $\operatorname{Re} s > -1$, so the difference on the left-hand side can be analytically continued to this region. Consequently, the analytic continuation $\zeta(s \mid a,b)$ of your sum satisfies

$$ \zeta(s \mid a,b) = \zeta(s+1,a+1) + g(s) $$

for $\operatorname{Re} s > -1$, $s \neq 0$ and thus has a simple pole at $s=0$ with residue $1$.

For $|s| < 1$ we have

$$ g(s) = \sum_{k=0}^\infty \left( \sum_{n=1}^\infty \frac{[\log(n+a)]^k}{(n+a)(n+b)} \right) \frac{(-s)^k}{k!}, $$

so here

$$ \zeta(s \mid a,b) = \frac{1}{s} + \sum_{k=0}^{\infty} \left( \gamma_{k}(a+1) + \sum_{n=1}^\infty \frac{[\log(n+a)]^k}{(n+a)(n+b)} \right) \frac{(-s)^k}{k!}. $$

For $s > 0$ we have

$$ \sum_{n=1}^{\infty} \frac{1}{(n+a)^s(n+b)} - \sum_{n=1}^{\infty} \frac{1}{(n+a)^{s+1}} = \sum_{n=1}^{\infty} \frac{a-b}{(n+a)^{s+1}(n+b)} =: g(s). $$

The series on the right-hand side converges and is analytic on $\operatorname{Re} s > -1$, so the difference on the left-hand side can be analytically continued to this region. Consequently, the analytic continuation $\zeta(s \mid a,b)$ of your sum satisfies

$$ \zeta(s \mid a,b) = \zeta(s+1,a+1) + g(s) $$

for $\operatorname{Re} s > -1$, $s \neq 0$ and thus has a simple pole at $s=0$ with residue $1$.

For $|s| < 1$ we have

$$ g(s) = \sum_{k=0}^\infty \left( \sum_{n=1}^\infty \frac{[\log(n+a)]^k}{(n+a)(n+b)} \right) \frac{(-s)^k}{k!}, $$

so for $0 < |s| < 1$

$$ \zeta(s \mid a,b) = \frac{1}{s} + \sum_{k=0}^{\infty} \left( \gamma_{k}(a+1) + \sum_{n=1}^\infty \frac{[\log(n+a)]^k}{(n+a)(n+b)} \right) \frac{(-s)^k}{k!}. $$