I was looking at a slight variation of the Möbius inversion and tried to prove the following direction:
If $$ g(n) = \sum_{\substack{d \in \mathcal{D}\\n | d}}{\mu\left(\frac{d}{n}\right) f(d)}$$ for all $n \in \mathcal{D}$, then $$f(n) = \sum_{\substack{d\in\mathcal{D}\\ n|d}}{g(d)}$$ for all $n \in \mathcal{D}$, with $\mathcal{D}$ being a finite divisor-closed set and $\mu$ the Möbius function.
I started with $$ \sum_{\substack{d\in\mathcal{D}\\n|d}}{g(d)} = \sum_{\substack{d\in\mathcal{D}\\n|d}}{\sum_{\substack{k\in\mathcal{D}\\d|k}}{\mu\left(\frac{k}{d}\right)f(k)}} = \sum_{\substack{d\in\mathcal{D}\\n|d}}{\sum_{dm\in\mathcal{D}}{\mu(m)f(dm)}}$$ Now I tried using $n \cdot l = d$ but seemed to get nowhere. How do I get on and get to the result which I guess looks something like $$ \sum_{nr \in \mathcal{D}}{\sum_{m|r}{f(nr)\mu(m)}}$$
If $S$ is a set of prime powers (of different primes) and $e^S$ the set of integers it generates then
$$\prod_{p^k \in S}(1-p^{-sk}) = \sum_{n \in e^D} \mu(n)n^{-s}, \qquad \qquad \prod_{p^k \in S}\frac{1}{1-p^{-sk}}= \sum_{n \in e^D} n^{-s}$$ $$ \sum_{n=1}^\infty n^{-s}\sum_{d | n,d \in e^S} \mu(d) f(n/d) = \sum_{n=1}^\infty f(n) n^{-s}\prod_{p^k \in S}(1-p^{-sk}) $$ $$\implies \sum_{n=1}^\infty n^{-s}\sum_{k | n,k \in e^S}\sum_{d | k,d \in e^S} \mu(d) f(k/d) = \prod_{p^k \in S}(1-p^{-sk})(\sum_{n=1}^\infty f(n) n^{-s}) \prod_{p^k \in S} \frac{1}{1-p^{-sk}} = \sum_{n=1}^\infty f(n) n^{-s}$$
$$\implies \sum_{k | n,k \in e^S}\sum_{d \in e^S} \mu(d) f(k/d) = f(n)$$