<s>[INST] Given the Dirichlet generating function for the divisor sum case, what is the Dirichlet generating function for the Möbius function case? From a search in the OEIS I found from the comment by Antonio G. Astudillo that:
$$\sum _{n=1}^{\infty} \frac{\left(\sum\limits_{d|n}d^{c}\right) \left(\sum\limits_{d|n}d^{z}\right)}{n^{\text{s}}}=\frac{\zeta (s) \zeta (s-c) \zeta (s-z) \zeta (-c+s-z)}{\zeta (-c+2 s-z)}$$
What is then the Dirichlet generating function for:
$$\sum _{n=1}^{\infty} \frac{\left(\sum\limits_{d|n}\mu(d) d^{c}\right) \left(\sum\limits_{d|n}\mu(d)d^{z}\right)}{n^{\text{s}}}=?$$
where $\mu(d)$ is the Möbius function of a divisor $d$. Above $\zeta(s)$ is of course the Riemann zeta function.
 [/INST] First let us try to evaluate
$$L_1(s) = \sum_{n\ge 1} \frac{1}{n^s} q_{a,b}(n) =
\sum_{n\ge 1} \frac{1}{n^s}
\left(\sum_{d|n} d^a\right) \left(\sum_{d|n} d^b\right).$$
With  the prime factorization of $n$ being
$$n = \prod_p p^v$$
we get for 
$$q_{a,b}(n) = 
\prod_p \frac{p^{av+a}-1}{p^a-1}
\frac{p^{bv+b}-1}{p^b-1}.$$
This yields for the Euler product
$$L_1(s) = \prod_p 
\left(1 + \sum_{v\ge 1} \frac{1}{p^{vs}}
\frac{p^{av+a}-1}{p^a-1}
\frac{p^{bv+b}-1}{p^b-1}\right)
\\ = \prod_p 
\left(1 + \frac{1}{p^a-1} \frac{1}{p^b-1}
\sum_{v\ge 1} \frac{1}{p^{vs}}
(p^{(a+b)v+a+b} - p^{av+a} - p^{bv+b} + 1)\right)
\\ = \prod_p 
\left(\frac{1}{p^a-1} \frac{1}{p^b-1}
\sum_{v\ge 0} \frac{1}{p^{vs}}
(p^{(a+b)v+a+b} - p^{av+a} - p^{bv+b} + 1)\right)
\\ = \prod_p 
\left(\frac{1}{p^a-1} \frac{1}{p^b-1}
\\ \times 
\left(\frac{p^{a+b}}{1-1/p^{s-(a+b)}}
- \frac{p^{a}}{1-1/p^{s-a}}
- \frac{p^{b}}{1-1/p^{s-b}}
+ \frac{1}{1-1/p^s}
\right)\right)
\\ = \prod_p 
\left(\frac{1}{p^a-1} \frac{1}{p^b-1}
\frac{1}{1-1/p^{s-(a+b)}}
\frac{1}{1-1/p^{s-a}}
\frac{1}{1-1/p^{s-b}}
\frac{1}{1-1/p^s}
\\ \times \left(p^{a+b}-p^a-p^b+1
- (p^{2a+2b} - p^{2a+b} - p^{a+2b} + p^{a+b})/p^{2s}
\right)\right)
\\ = \prod_p 
\left(
\frac{1}{1-1/p^{s-(a+b)}}
\frac{1}{1-1/p^{s-a}}
\frac{1}{1-1/p^{s-b}}
\frac{1}{1-1/p^s}
\left(1-p^{a+b}/p^{2s}\right)\right).$$
We now obtain by inspection that
$$L_1(s) = \frac{\zeta(s-(a+b))\zeta(s-a)\zeta(s-b)\zeta(s)}
{\zeta(2s-(a+b))}$$
as claimed. Continuing we introduce
$$L_2(s) = \sum_{n\ge 1} \frac{1}{n^s} r_{a,b}(n) =
\sum_{n\ge 1} \frac{1}{n^s}
\left(\sum_{d|n} \mu(d) d^a\right) 
\left(\sum_{d|n} \mu(d) d^b\right).$$
This time we have
$$r_{a,b}(n) = 
\prod_p (1-p^a) (1-p^b).$$
We get the Euler product
$$L_2(s) = \prod_p 
\left(1 + \sum_{v\ge 1} \frac{1}{p^{vs}}
(1-p^a)(1-p^b)\right)
\\= \prod_p 
\left(1 + (1-p^a)(1-p^b) \frac{1/p^s}{1-1/p^s}\right)
\\ = \zeta(s)
\prod_p (1 - 1/p^s + 
1/p^s - 1/p^{s-a} - 1/p^{s-b} + 1/p^{s-(a+b)})
\\ = \zeta(s)
\prod_p \left(1 + \frac{p^{a+b}-p^a-p^b}{p^s}\right)
\\ = \zeta(s)
\sum_{n\ge 1} \frac{1}{n^s} \mu(n)^2
\prod_{p|n} (p^{a+b}-p^a-p^b).$$
The inner product runs  over genuine prime divisors excluding one.
This seems to be the best we  can do at this point. What we have shown
here is that
$$\sum_{d|n} \mu(d)^2 \prod_{p|d} (p^{a+b}-p^a-p^b)
= \left(\sum_{d|n} \mu(d) d^a\right) 
\left(\sum_{d|n} \mu(d) d^b\right).$$
</s>