Is there a nice generating function proof of the following identity?

Question

Consider the Jordan function $J_2(n)$ defined by $$ J_2(n) = \#\{x \in (\mathbb{Z}/n)^2 \mid ord(x) = n\} $$ (this is OEIS A007434). One can prove the following identity pretty easily: $$ \sum_{d \mid n} J_2(d) \sigma_1(n/d) = n\sigma_1(n) $$ by noting that both sides are multiplicative functions, and so it suffices to check that this holds for $n = p^k$.

One could alternatively look at the Dirichlet generating functions; specifically, let $$ F(s) = \sum_{n=1}^\infty \frac{\sigma_1(n)}{n^s} $$ and $$ G(s) = \sum_{n=1}^\infty \frac{J_2(n)}{n^s} $$ in which case our identity is equivalent to the identity $$ F(s)G(s) = F(s-1). $$ Is there a nice generating function argument for this? This seems like a much nicer statement than hands-on computation, but I don't really know how to prove it.

I should remark that $G(s) = \frac{\zeta(s-2)}{\zeta(s)}$, so this is equivalent to the identity $$ F(s)\zeta(s-2) = F(s-1)\zeta(s). $$

Answer 1

The identity $G(s)=\frac{\zeta(s-2)}{\zeta(s)}$ is a corollary of the formula $J_2(p^e) = p^{2e} - p^{2e-2}$. The last identity from the question is simple because $F(s)=\zeta(s)\zeta(s−1).$

Answer 2

The identity $G(s)\zeta(s) = \zeta(s-2)$ has an easy direct proof: We are being asked to show that $$\sum_{d \geq 1} \frac{J_2(d)}{d^s} \sum_{e \geq 1} \frac{1}{e^s} = \sum_{n \geq 1} \frac{n^2}{n^s}$$ or, in other words, $$\sum_{d|n} J_2(d) = n^2.$$ The right hand side is the number of elements of $(\mathbb{Z}/n \mathbb{Z})^2$; the left hand side is the number of those elements with order $d$.

I don't see a good combinatorial interpretation of the fact that $\frac{\zeta(s-2)}{\zeta(s)} = \frac{\zeta(s-2)}{\zeta(s-1)} \frac{\zeta(s-1)}{\zeta(s)}$.