Conditions under which $\lim_{s\to1^+}\sum_{n=1}^{\infty}\frac{a_n}{n^s}=\sum_{n=1}^{\infty}\frac{a_n}{n}$

Here are analogous Tauberian theorems for power series and Dirichlet series that involve a condition of analytic continuation to a boundary point plus one extra condition that is necessary for the series to converge at some point on that boundary.

Power series: If $c_n \to 0$ then $\sum_{n \geq 0} c_nz^n$ converges for $|z| < 1$. Fatou showed this series converges at each number $z$ with $|z| = 1$ to which the series admits an analytic continuation from inside the disc $\{z : |z| < 1\}$. Note that "$a_n \to 0"$ is a necessary condition for $\sum c_nz^n$ to converge at a number $z$ where $|z| = 1$, so including it as a hypothesis is not unreasonable.

Dirichlet series: If $(a_1 + \cdots + a_n)/n \to 0$ then $\sum a_n/n^s$ converges for ${\rm Re}(s) > 1$. Riesz showed this series converges at each $s$ with ${\rm Re}(s) = 1$ to which the series admits an analytic continuation from the half-plane ${\rm Re}(s) > 1$. The condition $(a_1 + \cdots + a_n)/n \to 0$ is necessary for $\sum a_n/n^s$ to converge at a number $s$ where ${\rm Re}(s) = 1$, so including that as a hypothesis is not unreasonable.

For your motivating example with $a_n = \mu(n)$, of course the hypothesis $(a_1 + \cdots + a_n)/n \to 0$ is known by completely elementary methods to be equivalent to the Prime Number Theorem (in the form "$M(x) = o(x)$" for $M(x) = \sum_{m \leq x} \mu(m)$), so this theorem of Riesz isn't really a good approach to proving the Prime Number Theorem even though the analytic continuation hypothesis for $\sum \mu(n)/n^s = 1/\zeta(s)$ from ${\rm Re}(s) > 1$ to ${\rm Re}(s) = 1$ is valid.

In D. J. Newman's paper

A simple analytic proof of the prime number theorem

published in The American Mathematical Monthly 87 (1980) 693-696, Newman proved a result of this type which may also be useful to you (this is a minor variant of the criterion given by KConrad in his answer).

It says the following:

Let $|a_n| \leq 1$ and suppose that the Dirichlet series

$$\sum_{n = 1}^\infty \frac{a_n}{n^s}$$

admits a holomorphic continuation to the line $\mathrm{re}\, s = 1$. Then

$$\sum_{n = 1}^\infty \frac{a_n}{n^s}$$ converges for all $\mathrm{re}\, s \geq 1$.