Analytic continuation of Dirichlet function

We have $f(z) - b\zeta(z) = \sum_{n=1}^\infty \frac{c_n}{n^z}$ where $c_n=a_n-b$. The assumption gives you $|c_1+\cdots+c_n| \le C n^\sigma$. You should have a theorem available to you that tells you that under that condition, $\sum_{n=1}^\infty \frac{c_n}{n^z}$ converges for all $z$ with $\Re z>\sigma$, which would solve your problem.

To do it by hand, let $T(n) = c_1 + \cdots + c_n$. Then for real $x>\sigma$, $$ \sum_{n=1}^\infty \frac{c_n}{n^x} = \sum_{n=1}^\infty \frac{T(n)-T(n-1)}{n^x} = \sum_{n=1}^\infty T(n) \bigg( \frac1{n^x} - \frac1{(n+1)^x} \bigg). $$ By the mean value theorem, $$ \frac1{n^x} - \frac1{(n+1)^x} = (n-(n-1))\frac{-x}{\eta^{x+1}} = \frac x{\eta^{x+1}} $$ for some $\eta\in [n,n+1]$; in particular, it is at most $x/n^{x+1}$. Therefore $$ \bigg| \sum_{n=1}^\infty \frac{c_n}{n^x} \bigg| \le \sum_{n=1}^\infty Cn^\sigma \frac x{n^{x+1}} = Cx \sum_{n=1}^\infty \frac1{n^{x+1-\sigma}}, $$ which converges when $x\ge\sigma$. This proves the desired statement when $z$ is real, but you should have access to a theorem that deduces the full statement (convergence in the right-half plane) from that.