Extending the prime zeta function to negative values

Question

Consider the prime zeta function \begin{align} P(s) &= \sum_{p \text{ prime}} \frac{1}{p^s} \\ &= \sum_{n > 0} \frac{\mu(n)}{n} \log \zeta(n s) \end{align}

It has a singularity at the reciprocal of every squarefree positive integer $n$, since $\mu(n) \neq 0$ and $ns = 1$, where $\zeta$ has a pole. Noting that $\lim_{s \rightarrow 1} \zeta(s)(s-1) = 1$, let

\begin{align} f(s) &= P(s) + \sum_{n > 0} \frac{\mu(n)}{n} \log (ns - 1) \\ &= \sum_{n > 0} \frac{\mu(n)}{n} \log [\zeta(n s) (n s - 1)] \\ &= \sum_{\substack{n > 0 \\ n \neq 1/s}} \frac{\mu(n)}{n} \log [\zeta(n s) (n s - 1)] \\ \end{align}

Its graph looks like

f[s_, max_] := Sum[MoebiusMu[n]/n Log[
    If[n s == 1, 1, Zeta[n s] (n s - 1)]
    ], {n, 1, max}]
Plot[{
  f[s, 1000],
  f[s, 5000]
  }, {s, 0, 2}]

In particular, it seems smooth for all $s > 0$. Can the result be smoothly extended to $s < 0$? If so, is there a limit or convergent series describing this extension? The concepts of Taylor series, Mittag-Leffler star, and monodromy theorem might be relevant.

Edit: Let $g(s) = \log \zeta(s)(s-1)$. Then \begin{align} f(s) &= \sum_{n>0} \frac{\mu(n)}{n} g(ns) \\ f^{(1)}(s) &= \sum_{n>0} \frac{\mu(n)}{n} \frac{\partial}{\partial s} g(ns) \\ &= \sum_{n>0} \frac{\mu(n)}{n} \frac{\partial ns}{\partial s} \frac{\partial}{\partial ns} g(ns) \\ &= \sum_{n>0} \mu(n) g^{(1)}(ns) \\ f^{(2)}(s) &= \sum_{n>0} \mu(n) \frac{\partial}{\partial s} g^{(1)}(ns) \\ &= \sum_{n>0} \mu(n) n g^{(2)}(ns) \\ f^{(k)}(s) &= \sum_{n>0} \mu(n) n^{k-1} g^{(k)}(ns) \\ \end{align}

\begin{align} f^{(k)}(0) &= \sum_{n>0} \mu(n) n^{k-1} g^{(k)}(0) \\ &= g^{(k)}(0) \sum_{n>0} \frac{\mu(n)}{n^{1-k}} \\ &= \frac{g^{(k)}(0)}{\zeta(1-k)} \\ &= \frac{(-1)^{k+1} k}{B_k} g^{(k)}(0) \\ f^{(0)}(0) &= 0 \\ f^{(1)}(0) &= 2 - 2 \log 2\pi \\ &\approx -1.67575 \\ f^{(2)}(0) &= 6 + 6 \gamma_0^2 - \frac{\pi^2}{2} + 12 \gamma_1 \\ &\approx 2.19048 \end{align}

However, $B_k = 0$ if $k$ is odd and $k \geq 3$. Does this mean $f^{(k)}(0)$ is undefined for those values? We could potentially regularize $\zeta(1-k)^{-1}$ at those $k$ by taking the Cauchy principal values:

\begin{align} \zeta(1-k)^{-1} &\overset{\star}{=} \lim_{s \rightarrow 0} \frac{\zeta(1-k+s)^{-1} + \zeta(1-k-s)^{-1}}{2} \\ &= -\frac{\zeta''(1-k)}{2 \zeta'(1-k)^2} \end{align}