Does this product have analytic continuation?

Let $\sigma:=\mathrm{Re}(s)$, and consider the principal branch of the logarithm. For $\sigma>3/2$ we have $$\begin{align*}p\log(1-p^{-s})-\log(1-p^{1-s})&=p\left(-p^{-s}+O(p^{-2\sigma})\right)-\left(-p^{1-s}+O(p^{2-2\sigma})\right)\\&=O(p^{2-2\sigma}),\end{align*}$$ hence the "Euler sum" $$H(s):=\sum_p\left\{p\log(1-p^{-s})-\log(1-p^{1-s})\right\},\qquad\sigma>3/2,$$ converges locally uniformly (and absolutely). This implies that the Euler product $$ G(s):=\exp(H(s))=\prod_p\frac{(1-p^{-s})^p}{1-p^{1-s}},\qquad\sigma>3/2,$$ defines a non-vanishing holomorphic function. In the original half-plane $\sigma>2$, we have $$ G(s)=\frac{\zeta(s-1)}{F(s)},\qquad\sigma>2,$$ hence $F(s)=\zeta(s-1)/G(s)$ extends to a meromorphic function in $\sigma>3/2$ with a simple pole at $s=2$ and no other pole there.

Regarding your second question, I am not aware of any papers where this function was studied. My argument above is rather standard though.

$$P(s) = \sum_p p^{-s}, \qquad \log F(s) = \sum_{p^k} \frac{p^{1-sk}}{k} = \sum_{k\ge 1} \frac{P(sk-1)}{k}$$

• $P(s) = \sum_{n=1}^\infty \frac{\mu(n)}{n} \log \zeta(ns)$ and $P_N(s) = \sum_{n=N+1}^\infty \frac{\mu(n)}{n} \log \zeta(ns)$ is analytic for $\Re(s) > \frac{1}{N+1}$ so that $$e^{N! P(s)} = e^{N! P_N(s)}\prod_{n=1}^{N-1} \zeta(ns)^{\mu(n) \frac{N! }{n}}$$ is meromorphic for $\Re(s) > \frac{1}{N+1}$ providing the analytic continuation of $P(s)$ :

  $P(s)$ has a branch point at $\frac{\rho}{N}$ for each $N\ge 1$ and non-trivial zero $\rho$ of $\zeta$.

  Therefore $P(s)$ has a natural boundary on $\Re(s) = 0$ and no analytic continuation exists beyond there.

• For the same reason $F(s)^{N!}$ is meromorphic for $\Re(s) > 1+\frac{1}{N+1}$ and $\log F(s)$ has a branch point at $1+\frac{\rho}{N}$ for every $N\ge 1,\rho$ and hence a natural boundary on $\Re(s) = 1$ and no analytic continuation exists beyond there.