We can solve this one using species which possibly qualifies as a combinatorial proof.
First, observe that 2-regular graphs are sets of undirected cycles, which is the species $\mathfrak{P}( \mathfrak{D}_{=3}(\mathcal{Z}) + \mathfrak{D}_{=4}(\mathcal{Z}) + \mathfrak{D}_{=5}(\mathcal{Z}) + \cdots).$
This gives the generating function $G(z) = \exp \left(\frac{1}{2} \frac{z^3}{3} + \frac{1}{2} \frac{z^4}{4} + \frac{1}{2} \frac{z^5}{5} + \frac{1}{2} \frac{z^6}{6} + \cdots\right),$ which is $G(z) = \exp\left(-\frac{z}{2}-\frac{z^2}{4}\right) \frac{1}{\sqrt{1-z}}.$
On the other hand, permutations with no fixed points or 2-cycles are the species $\mathfrak{P}( \mathfrak{C}_{=3}(\mathcal{Z}) + \mathfrak{C}_{=4}(\mathcal{Z}) + \mathfrak{C}_{=5}(\mathcal{Z}) + \cdots)$
which gives the generating function $C(z) = \exp \left(\frac{z^3}{3} + \frac{z^4}{4} + \frac{z^5}{5} + \frac{z^6}{6} + \cdots\right),$ which is $C(z) = \exp\left(-\frac{z}{1}-\frac{z^2}{2}\right) \frac{1}{1-z}.$
Observe that when we multiply two exponential generating functions of the sequences $\{a_n\}$ and $\{b_n\}$ we get that
$ A(z) B(z) = \sum_{n\ge 0} a_n \frac{z^n}{n!} \sum_{n\ge 0} b_n \frac{z^n}{n!} = \sum_{n\ge 0} \sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} a_k b_{n-k} z^n\\ = \sum_{n\ge 0} \sum_{k=0}^n \frac{n!}{k!(n-k)!} a_k b_{n-k} \frac{z^n}{n!} = \sum_{n\ge 0} \left(\sum_{k=0}^n {n\choose k} a_k b_{n-k}\right)\frac{z^n}{n!}$
i.e. the product of the two generating functions is the generating function of $\sum_{k=0}^n {n\choose k} a_k b_{n-k}.$
In the present case we clearly have $A(z) = B(z) = G(z)$ and we need to prove that their convolution is $C(z).$
Note however that $G(z)^2 = \left(\exp\left(-\frac{z}{2}-\frac{z^2}{4}\right)\right)^2 \frac{1}{\sqrt{1-z}^2}$ which is $\exp\left(-\frac{z}{1}-\frac{z^2}{2}\right) \frac{1}{1-z},$ i.e. precisely $C(z),$ QED.
Here we have used the fact that the dihedral group acting on $q$ elements contains $2q$ permutations and the cyclic group $q$ permutations.