Well, a simple counter example is $$A(s)=\sum_{n=1}^\infty a_n n^{-s}= e^{\eta(s)},$$
where
$$\eta(s)=\sum_{n=1}^\infty (-1)^{n-1} n^{-s}=(1-2^{1-s})\zeta(s).$$
This Dirichlet series is obviously meromorphic since it is in fact entire and it is also absolutely convergent on some half plane Re$(s)>c$. This entire function is not of finite order by the functional equation of the Riemann zeta-function and Stirling's formula.

Update: An even simpler example along the same lines is
$$B(s)=\sum_{k=0}^\infty \frac{2^{-ks}}{k!} =e^{2^{-s}}.$$
This Dirichlet series is an entire function that is absolutely convergent in the full complex plane, so it is absolutely convergent for any half plane Re$(s)>c$. However we have that $B(-x)= e^{2^x}$, and thus it does not fulfill $B(-x)\ll e^{x^c}$ for any $c>0$ and it is not an entire function of finite order. If we would like to have a meromorphic function with some pole that has some abscissa of convergence we can consider $C(s)=\zeta(s)+B(s)$. This function is an ordinary Dirichlet series that is absolutely convergent if and only if Re$(s)>1$, has a pole at Re$(s)=1$ and meromorphic continuation to the entire complex plane, but it does not have finite order.