Let $\tau$ denote the classical divisor function and $\mu$ be the Möbius function.
Then for each pair of integers $n,m$ we have $$\tau(mn)=\sum_{d\mid(m,n)}\mu(d) \tau(m/d)\tau(n/d),$$ where the sum is taken over all positive integer common divisors of $m$ and $n$. I can verify this by using multiplicativity and checking via brute force that it is true when $m,n$ are powers of the same prime.
My question is whether there is a different proof and whether it is part of a bigger family of similar identities.
We seek to show that
$$\tau(mn) = \sum_{d|(m,n)} \mu(d)\tau(m/d)\tau(n/d).$$
Suppose first that there are prime factors not shared between $m$ and $n.$ We show that we can in fact assume that there are no such factors. Let $m'|m$ and $n'|n$ be the maximal contribution from the shared factors so that for a prime $p|(m,n)$ it does not divide $m/m'$ and $n/n'$.By virtue of multiplicativity we get for the LHS
$$\tau(m')\tau(m/m') \tau(n') \tau(n/n')$$
and for the RHS
$$\sum_{d|(m,n)} \mu(d) \tau(m'/d) \tau(m/m') \tau(n'/d) \tau(n/n').$$
We see that the product of the not shared component $\tau(m/m')\tau(n/n')$ cancels. Therefore we may assume that $m$ and $n$ have the same set of prime factors. Let these primes be indexed as $p_j$, with the exponent of $p_j$ in $m$ being $v_j$ and in $n$ being $w_j.$ We have for the sum on the RHS the product representation (these are just the definitions of $\tau$ and $\mu$)
$$\tau(m)\tau(n)\prod_{j} \left(1+\mu(p_j) \frac{\tau(p_j^{v_j-1})\tau(p_j^{w_j-1})} {\tau(p_j^{v_j})\tau(p_j^{w_j})}\right) \\ = \tau(m)\tau(n)\prod_{j} \left(1-\frac{v_j w_j}{(v_j+1)(w_j+1)}\right) \\ = \tau(m)\tau(n) \prod_{j} \frac{v_j+ w_j+1}{(v_j+1)(w_j+1)} \\ = \tau(mn)\prod_j \frac{(v_j+1)(w_j+1)}{v_j+w_j+1} \prod_{j} \frac{v_j+ w_j+1}{(v_j+1)(w_j+1)} \\ = \tau(mn).$$
This is the claim. (On the first line we have a factor of $-1$ for $\mu(d)$ and we must replace $v_j+1$ and $w_j+1$ in the products of $\tau(m)$ and $\tau(n)$ by $v_j$ and $w_j$ to obtain $\tau(m/d)$ and $\tau(n/d).$)