Let $F(s)$ be a Selberg function, $F(s)=\sum_{n=1}^\infty\frac{a(n)}{n^{s}}$ and $\log F(s)=\sum_{n=2}^\infty\frac{b(n)\Lambda(n)}{\log n}\frac{1}{n^s}$ where $b(n)$ is zero unless $n$ is a positive power of a prime, i.e. $F(s)=\prod_p\exp\left(\sum_{k=1}^\infty\frac{b(p^k)}{kp^{ks}}\right)=:\prod_p F_p(s)$. It is well known that then the $a(n)$ are multiplicative, but I couldn't find a proof yet. Of course, it suffices to show that $F_p(s)=\sum_{k=0}^\infty\frac{a(p^k)}{p^{ks}}$.
So far I got this: Define $G(s):=\sum_{n=2}^\infty\frac{b(n)\Lambda(n)}{\log n}\frac{1}{n^s}$, so we have $F(s)=e^{G(s)}$ with $G(s)$ holomorphic in a half plane. So, by the uniqueness of the complex logarithm we find that $G(s)=\log a(1)+2\pi il+\sum_{n=2}^\infty\frac{(a'\ast a^{-1})(n)}{\log n}\frac{1}{n^s}$ (where $a'(n)=a(n)\log n$), so comparing with the definition of $G(s)$ and with $a(1)=1$ we get $b(n)\Lambda(n)=(a'\ast a^{-1})(n)$ for all $n\geq2$.
Repeating this argumentation for $F_p(s)$ where $p$ is fixed we see, that if $F_p(s)=\sum_{n=1}^\infty\frac{c(n)}{n^s}$ for some $c(n)$, then $b(p^k)=\frac{(c'\ast c^{-1})(p^k)}{\log p}$ for every $k\geq 1$, $c(1)=1$ and $\forall n>1:(c'\ast c^{-1})(n)=0$ unless $n$ is a positiv power of $p$.
The next natural step would be to expand $F_p(s)$ into a Dirichlet series using the power series of $\exp$. I think, if I knew $F_p(s)$ was an absolutely convergent Dirichlet series for all $p$, I could do the proof.
So what's next? I hope anybody has an idea.