Yes. All the $r_i$ must equal $0$ if the period is prime, however. Consider for example $$f(s)=(1-p^{1-s})^2 \zeta(s),$$ which is periodic with period $p^2$, at $s=1$.

I should probably expand on this answer a bit. The case where $t$ is prime is an old conjecture of Chowla, which was resolved by Baker, Birch, and Wirsing (all the $r_i=0$ in this case) in the paper I link to in the first word of this answer. They give the Dirichlet series for $f(s)$ above as a counterexample when $t$ is not prime.

To see that $f(s)$ has the desired properties, I'll work it out in a bit more detail for $p=2$. Expanding $f$ out as a Dirichlet series gives $$f(s)=\sum_{n=0}^\infty \frac{1}{(4n+1)^s}-\frac{3}{(4n+2)^s}+\frac{1}{(4n+3)^s}+\frac{1}{(4n+4)^s}$$ as Woett remarks in the comments. On the other hand, $(1-2^{1-s})^2$ has a double zero at $s=1$, whereas the zeta function $\zeta(s)$ has a simple pole at $s=1$; so $f(1)=0$. So taking the limit as $s\to 1^+$ gives that the OP's series converges to $f(1)=0$ for $r_1=r_3=r_4=1,~ r_2=-3, ~t=4$, as desired.