Let us define an Arithmetical function $\nu(1)=0$. For $n > 1$, let $\nu(n)$ be the number of distinct prime factors of $n$.
I need to prove $\mu * \nu (n)$ is always 0 or 1.
According to my computation, if $n$ is prime, it is 1. If $n$ is composite, it is 0.
I tried using induction, so that assuming for a composite number $m$, $\mu*\nu(m) = 0$, then for any prime $q$, $\mu*\nu(qm) = 0$.
On
$$F(a,s) = \prod_p (1+a \sum_{k=1}^\infty p^{-sk}) = \sum_{n=1}^\infty n^{-s} a^{\nu(n)}$$
$$\frac{\partial F(a,s)}{\partial a}|_{a=1} = \sum_{n=1}^\infty n^{-s} \nu(n)$$
$$G(a,s) = \frac{F(a,s)}{\zeta(s)} = \prod_p (1-p^{-s}) (1+a \sum_{a=1}^\infty p^{-sk}) = \prod_p (1+(a-1)p ^{-s}) $$
$$\frac{\partial G(a,s)}{\partial a}|_{a=1} = G(1,s)\sum_p \frac{\partial (1+(a-1)p ^{-s})}{\partial a}|_{a=1} = G(1,s) \sum_p p^{-s} = \sum_p p^{-s}$$
thus $\nu \ast \mu (n) = \delta_\pi(n)$ where $\delta_\pi(n)$ is the prime indicating function
Start from the Euler product
$$F(a,s) = \prod_p \left(1 + \frac{a}{p^s} + \frac{a}{p^{2s}} + \cdots\right)$$
which by inspection evaluates to
$$F(a, s) = \sum_{n\ge 0} \frac{a^{\nu(n)}}{n^s}.$$
This has the property that
$$H(a, s) = \left.\frac{\partial}{\partial a} F(a,s)\right|_{a=1} = \sum_{n\ge 1} \frac{\nu(n)}{n^s}.$$
On the other hand the derivative of $F(a,s)$ is given by $$\prod_p \left(1 + a\frac{1/p^s}{1-1/p^s}\right) \sum_p \left(1 + a\frac{1/p^s}{1-1/p^s}\right)^{-1} \frac{1/p^s}{1-1/p^s}.$$
Evaluate this at $a=1$ to get $$H(a, s) = \zeta(s) \sum_p \left(\frac{1}{1-1/p^s}\right)^{-1} \frac{1/p^s}{1-1/p^s} = \zeta(s) \sum_p \frac{1}{p^s}.$$
Now for the convolution $$\mu * \nu = \sum_{d|n} \mu(d)\nu(n/d)$$ we use the Dirichlet series
$$L(s) = \sum_{n\ge 1} \frac{\mu(n)}{n^s} = \frac{1}{\zeta(s)}$$
to obtain $$L(s) H(a, s) = \frac{1}{\zeta(s)} \zeta(s) \sum_p \frac{1}{p^s} = \sum_p \frac{1}{p^s}$$
and we have the prime indicator as claimed.