Convergence of Euler product implies convergence of Dirichlet series?

(Crossposted to Math Overflow) Suppose we have an Euler product over the primes

$$F(s) = prod_{p} left( 1 – frac{a_p}{p^s} right)^{-1},$$

where each $a_p in mathbb{C}$. The Euler product is convergent in the range $Re(s) > sigma_c$, and absolutely convergent in the range $Re(s) > sigma_a$, for some $sigma_c < sigma_a in mathbb{R}$. If we multiply out the Euler product, we get a Dirichlet series

$$F(s) = sum_{n=1}^infty frac{a_n}{n^s},$$

where $a_n = prod_{p^k || n} a_p^k$ is completely multiplicative as a function of $n$.

Question: We know that the Dirichlet series for $F(s)$ must converge absolutely in the half-plane $Re(s) > sigma_a$. Must the Dirichlet series for $F(s)$ also converge in the half-plane $Re(s) > sigma_c$? If not, what is a counterexample?

Edit: My question is motivated by considering a product like

$$F(s) = left(1 – frac{1}{2^s}right)^{-1}left(1 + frac{1}{3^s}right)^{-1}left(1 – frac{1}{5^s}right)^{-1}left(1 + frac{1}{7^s}right)^{-1} … = prod_{n=1}^infty left( 1 + frac{(-1)^n}{p_n^s} right)^{-1},$$

where a classical result on infinite series demonstrates convergence for $Re(s) > 1/2$ [although absolute convergence only happens in the half-plane $Re(s) > 1$]. This product for $F(s)$ will have no zeroes in the half-plane $Re(s) > 1/2$, so if we multiply it out to get the Dirichlet series

$$F(s) = sum_{n=1}^infty frac{a_n}{n^s} = 1 + frac{1}{2^s} – frac{1}{3^s} + frac{1}{4^s} + frac{1}{5^s} – frac{1}{6^s} – frac{1}{7^s}…,$$

does the Dirichlet series converge too? Can we then conclude that the coefficients $a_n$ satisfy

$$sum_{j = 1}^n a_j = O(n^{1/2 + epsilon}),$$

for all $epsilon > 0$?