Can a Dirichlet series have infinitely many zeros and be nonzero?
To be precise, by a Dirichlet series I mean a function of the form $s\mapsto \sum_{n\geq 1}\frac{a_n}{n^s}$ where the domain is the half-plane of convergence of the series $\sum_{n\geq 1}\frac{a_n}{n^s}$ with $a_n\in\mathbb C$.
What I know: The zeros cannot accumulate nor can the real part of the zeros tend to infinity.
Yes. For a cheap set of examples, consider the everywhere-convergent Dirichlet series
$$ D_0(s) := 1 - \frac{1}{2^s}.$$
This has zeros at $s = 2 \pi i k / \log 2$ for any integer $k$ and converges everywhere.
We can construct far more sophisticated sets of examples too. Any Dirichlet series has the property that it is almost periodic in its domain of absolute convergence. This means that for any $\epsilon > 0$ and any $\alpha, \beta$ such that $s = \alpha + it$ and $s = \beta + it$ are within the region of absolute convergence, there exists a length $\ell$ such that every interval of length $\ell$ [notated generically here as $(t_1, t_2 = t_1 + \ell)$] contains an almost period; this means that there exists some $\tau \in (t_1, t_2)$ such that $$ \lvert f(\sigma + it + i\tau) - f(\sigma + it) \rvert < \epsilon \qquad \text{for any} \qquad \alpha \leq \sigma \leq \beta, t \in \mathbb{R}.$$ This is a cumbersome definition. In short, it means that on any line $\mathrm{Re} s = \sigma$ in the region of absolute convergence, the Dirichlet series is almost periodic (up to an error of $\epsilon$) with respect to at least one length $\tau$ on that line.
Bohr showed that every Dirichlet series is almost periodic in its half-plane of absolute convergence. More generally, Bohr showed that any function that can be uniformly approximated by trigonometric polynomials is, which descends to Dirichlet series. (Besicovitch's monograph Almost Periodic Functions has much more to say about this).
Proof: If $f$ is identically zero, we are done. So we now suppose $f$ is not identically zero.
Let $s_0$ be a zero in the strip. There exists some $\eta_0 > 0$ such that the circle $C = \{ s : \lvert s - s_0 \rvert = \eta_0 \}$ is entirely within the strip and such that $f$ has no zeros on $C$. (We are now using the assumption that $f$ is not identically zero).
Let $\varepsilon = \mathrm{inf}_{s \in C} \lvert f(s) \rvert$ be the minimum value of $f$ on the circle $C$. By almost periodicity, there exists a length $\ell$ such that any interval of length $\ell$ contains $\tau$ such that $$ \lvert f(s + i\tau) - f(s) \rvert \leq \varepsilon / 2, $$ in particular this holds for all $s \in C$. Rouche's Theorem then shows that $f$ has the same number of zeros (i.e. at least one) in $C$ and in $C + i\tau$.
The density then follows from almost periodicity, as at least one such $\tau$ exists in every interval of length $\ell$. And so in particular we should expect there to be at least $T / \ell$ zeroes with $0 \leq t \leq T$. $\square$
As Dirichlet series are almost periodic, this immediately gives the following corollary.
And in particular, these zeros all lie approximately on a vertical line (with epsilon fuzz). Thus finding a Dirichlet series with one zero in the region of absolute convergence is equivalent to finding a Dirichlet series with infinitely many zeros in the region of absolute convergence.
It follows that to construct examples of Dirichlet series with many zeros, you can combine other Dirichlet series. For example, take your two favorite (distinct) Dirichlet characters $\chi$ and $\psi$, and your preferred value $s_0$ with $\mathrm{Re}(s_0) > 1$. We know $L(s_0, \chi) := a_\chi \neq 0$ and $L(s_0, \psi) := a_\psi \neq 0$ because they have absolutely convergent Euler products. But then the Dirichlet series $$ D(s) := \frac{1}{a_\chi} L(s, \chi) - \frac{1}{a_\psi} L(s, \psi) $$ has $D(s_0) = 1 - 1 = 0$ by construction. Thus $D(s)$ has infinitely many zeros in the region of absolute convergence.