Let $P(s),C(s),\zeta(s)$ be the prime zeta function, the analogous composite zeta function, and the classical zeta function.
I do not know whether it is known that there are infinitely many zeros of C, (or P). As an exercise I tried to show this.
My question is whether A-C below is a plausible argument (in the sense that it could be made into a proof) for the proposition:
Proposition: There are infinitely many zeros of $C(s)$ (and $ P(s)$).
A. Loosely speaking, a zero of an analytic function $f$ is a continuous function of $f.$
Since $\zeta,P,C$ are analytic, it seems that Paul Rosenbloom's 1969 paper in J. of Approximation Thy, "Perturbation of the Zeros of Analytic Functions I", applies, in particular Hurwitz's thm. If $f_n\to f$ the limit pts. of the zeros of $f_n$ are the zeros of f. More to the point, he appears to show that (with conditions) if f is close to g, near any point where g is close to 0 there is a point where g is exactly zero.
B. We can map every nontrivial zero of $\zeta(s)$ to a zero or set of zeros of $C(s)$ and conversely. Starting at a nontrivial zero of $\zeta(s)$ we subtract (if necessary, small multiples of) terms of the analytic continuation of $P(s)$ from $\zeta(s),$ which shifts the zero of $\zeta$ stepwise towards that of $C(s).$ In a small finite number of steps the zero of the perturbed function approximates that of $C(s).$
C. Two zeros may map to the same zero of C (or P) but this is not I think an issue. If we can show that for given zero $z$ of $\zeta$ and the (finite) set of zeros $s$ to which $z$ can be perturbed, the largest distance $|z-s_o|$ is finite, then we only have to choose the next $z$--as $\Im(z)$ increases--sufficiently large that none of the zeros to which it perturbs could possibly be $s_o.$ Continuing in this way we get an infinite sequence of zeros of C.
One possible problem is that one may be drawing a zero of $P$ (or $C$) towards a point which is a zero of $\zeta,$ hence a badly-behaved point of (the analytic continuation of) P.
For example, it seems hard to find a path from $C(0.38+12.4~i)= 0$ to $\zeta(.5+14.1~i)=0,$ but by adding a convenient number $z$ we can get a path $C\to C+z \to \zeta+z \to \zeta$ which avoids discontinuities. Since we have a path from $C(.32+15.4~i)=0$ to $\zeta(0.5+14.1~i)=0$ the mapping is evidently many-to-one, but still would show infinitely many zeros of P and C if the idea is otherwise sound.
Example.
There is an analytic continuation of $\zeta$ for $\sigma \leq 1.$ There is an analytic continuation of $P$ valid for $ 0 < \sigma. $
The continuation for $P$ obtained by Mobius inversion is
$$P(s)= \sum_{k=1}^\infty \frac{\mu(k)}{k}\log \zeta(k s).$$
Let $P_N(s) = \sum_{k=1}^N \frac{\mu(k)}{k}\log \zeta (k s).$
$\zeta(1/2+56.446... i) = 0.$
$\zeta(.5+55.8~ i) - P_1(.5+55.8 ~i)\approx 0. $
$\zeta(.65+55.8 ~i)-P_2(.65+55.8 ~i)\approx 0.$
...
$\zeta(.64475 +55.8908~ i) - P_{20}(.64475+55.8908~ i) \approx C(.64475+55.8908 ~i)\approx 0.$
Detail of the above calculation with different zero: perturbing the first nontrivial zero of $\zeta$ to a zero of $\zeta(s)-(\mu(1)/1)\log \zeta(1(s))$ in increments of 1/20.
For the table the perturbation begins at the first nontrivial zero of $\zeta\approx (0.5+ 14.13~i).$
$\zeta(a+b~ i)- (m/20)(\mu(1)/1)\log \zeta(1(a+b~i))\approx 0$
$$\begin{array}{c | c | c | } m & a & b \\ \hline 0 & 0.5 & 14.13 \\ \hline 1 & 0.42 &14.3 \\ \hline 2 & 0.36& 14.4\\ \hline 3 & 0.33& 14.5 \\ \hline 4 & 0.32 & 14.6 \\ \hline ... & ... & ... \\ \hline 17& 0.24 & 15.4 \\ \hline 18 & 0.22 & 15.5\\ \hline 19 & 0.22 & 15.58\\ \hline 20 & 0.22 & 15.6 \\ \hline \end{array}$$
The analogous problem for $P(s)$ is intended as part of the question.