Suppose $(f_n(s))_n$ and $(g_n(s))_n$ are two sequences of Dirichlet series with positive coefficients such that $\exists \alpha\in\mathbb R$ such that for all $s\in\mathbb C$ with $\Re(s)>\alpha$ and for all $n$, the series $f_n(s)$ and $g_n(s)$ converge. Call the sequences $(f_n(s))_n$ and $(g_n(s))_n$ equivalent if there exists some $C>0$ such that for any $s\in\mathbb R$ such that $f_n(s)$ and $g_n(s)$ converge for all $n$ we have $$C^{-1-s}g_n(s)\le f_n(s)\le C^{1+s}g_n(s),\quad \forall n\in\mathbb N.$$
Put $F(s)=\prod_n(1+f_n(s))$ and $G(s)=\prod_n(1+g_n(s))$.
It is not too hard to check that if $(f_n(s))$ and $(g_n(s))$ are equivalent than for any $s\in\mathbb C$, $F(s)$ converges iff $G(s)$ converges.
My question is this: Suppose that $(f_n(s))$ and $(g_n(s))$ are equivalent and let $\alpha_0$ be the infimum $\alpha$ such that $F(s)$ (and hence $G(s)$ as well) converges on the half-plain $\Re(s)>\alpha$ (i.e. the abscissa of convergence of $F(s)$). Assume that there exists some $\delta>0$ such that $F(s)$ can be meromorphically continued to the half plain $\Re(s)>\alpha_0-\delta$.
Does it follow that $G(s)$ can be meromorphically continued to the half plain $\Re(s)>\alpha_0-\delta'$ for some $\delta'>0$?
Intuition: I have to say that my intuition says that the answer to this question is no, but I haven't managed to find a counter-example yet. The reason I believe it is false is since while the abscissa of convergence of a Dirichlet series is determined by the behavior of the series along the real-line- the question of meromorphic continuation is more subtle and require to understand the behavior of $F(s)$ along the line $\Re(s)=\alpha_0$.
But it is very possible that I am lacking knowledge on Dirichlet series an in fact such an equivalence does determine whether a function can be continued along the line $\Re(s)=\alpha_0$.
What I have done do far: So my attempts of finding a counter example reduce to trying to work with known sequences $(f_n(s))$ such that the product $F(s)=\prod_n(1+f_n(s))$ has a natural bound at imaginary line through the abscissa, and hence can not be continued to any half plain $\Re(s)>\alpha-\delta$.
One such example appears here, and is given by $f_p(s)=\frac{p^{-1-s}}{1-p^{-s}}=\sum_n p^{-1-(n+1)s}$, whenever $n=p$ is prime, and $f_n(s)=0$ otherwise. It can be checked that the product $F(s)=\prod_{p\:prime}1+\frac{p^{-1-s}}{1-p^{-s}}$ converges whenever $s>0$ and that $F(s)$ vanishes along the line $\Re(s)=0$ and hence can not be merom. continued past it.
It tried to an easy sequence $g_n(s)$ for which the product $G(s)=\prod_n(1+g_n(s))$ is known to have a merom. continuation past $0$ and such that $(g_n(s))$ and $f_n(s))$ are equivalent. For example $g_p(s)=\frac{p^{-1-s}}{1-p^{-1-s}}$ for $n=p$ prime and $g_n(s)=0$ otherwise, for which $G(s)=\zeta(s+1)$ is the shifted Riemann zeta function, and hence can be merom. continued. Unfortunately, for this $g_n$ it only holds that for any $s>0$ $$g_n(s)=g_n(s)\cdot 1^{-1-s}<f_n(s)$$ and there exists no $C>0$ such that $f_n(s)<C^{1+s}g_n(s)$ for all $n$'s. This follows simply from the fact that $f_n(s)/g_n(s)$ tends to infinity as $s\to 0$ for any $n$ prime.
In any case, if someone here has an idea about how to make this example into a counter-example, or maybe knows of another counter example, or otherwise knows of a reason why the answer to my question should actually be yes, I would appreciate a response very much.
Thank you, Shai
Remark- I did not write in any information for why this question interests me, as I thought this would be irrelevant. If anyone else thinks otherwise, I can definitely fill in this gap.