Question
Let $F(s) = \sum_{p^k} p^{-ks}$, where the summation runs over all prime powers. Show that $F(s) = \log \zeta(s) + G(s)$ holds for $\text{Re }s>1$, where $G(s)$ is analytic in the half-plane $\text{Re }s > 1/2$. Deduce from this that the function $F(s)$ does not have a meromorphic continuation for $\text{Re }s<1$.
My Attempt
I know that $$F(s) = \sum_{p} \sum_{k=1}^{\infty} \frac{1}{p^{ks}} = \sum_{p} \frac{p^{-s}}{1-p^{-s}}\, ,$$ and $$\log \zeta(s) = \log \left(\prod_{p} \left(1-\frac{1}{p^s}\right)^{-1}\right) = \sum_{p} \log \left(1-\frac{1}{p^s}\right)^{-1} = \sum_{p} \sum_{k=1}^{\infty} \frac{1}{k\, p^{ks}}.$$
But how to proceed naturally, instead of reverse engineering as I am trying to do here?
$$G(s)= \sum_p \sum_{k\ge 2} p^{-sk} (1-1/k)$$ converges absolutely (bounded by $2\zeta(2\Re(s))$) thus is analytic for $\Re(s) > 1/2$ and $$F(s) = \log \zeta(s)+G(s)= -\log (s-1)+ \log ((s-1)\zeta(s))+G(s)$$ has a branch point at $s=1$