Seeking combinatorial or group theoretic proof for permutation identity
Permutations with odd cycles only and cycle count marked have the species
$$\mathfrak{P}(\mathcal{U}\mathfrak{C}_{=1}(\mathcal{Z}) + \mathcal{U}\mathfrak{C}_{=3}(\mathcal{Z}) + \mathcal{U}\mathfrak{C}_{=5}(\mathcal{Z}) + \mathcal{U}\mathfrak{C}_{=7}(\mathcal{Z}) + \cdots).$$
which yields the EGF
$$G(z,u) = \exp\left(uz+ u\frac{z^3}{3}+ u\frac{z^5}{5}+ u\frac{z^7}{7}+\cdots\right) \\ = \exp\left(u\sum_{k\ge 0}\frac{z^{2k+1}}{2k+1}\right) \\ = \exp\left(u\log\frac{1}{1-z} - u\sum_{k\ge 1} \frac{z^{2k}}{2k}\right) \\ = \exp\left(u\log\frac{1}{1-z} - u\frac{1}{2}\log\frac{1}{1-z^2}\right).$$
Now here we have $u=4$ so we continue with
$$\frac{1}{(1-z)^4} \exp\left(2\log(1-z^2)\right) = \frac{(1-z^2)^2}{(1-z)^4} = \frac{(1+z)^2}{(1-z)^2} \\ = (1+2z+z^2) \frac{1}{(1-z)^2}.$$
We get as our result the quantity
$$\frac{1}{4} n! [z^n] G(z, 4) = \frac{1}{4} n! \left({n+1\choose 1} + 2{n\choose 1} + {n-1\choose 1}\right) \\ = \frac{1}{4} n! (n+1+2n+n-1) = n! \times n.$$
I am going to introduce a fancy structure, that is a coloured permutation.
A coloured permutation $\sigma$ is a permutation acting on $\{1,2,\ldots,n\}$, with the property that every element has an odd order. Moreover, the elements of $\{1,\ldots,n\}$ are coloured with some colour from $\{\text{cyan},\text{magenta},\text{yellow},\text{black}\}$ and the permutation is colour-preserving, i.e. the colour of $\sigma(a)$ is the same as the colour of $a$. We want to count the number of coloured permutations, given by: $$ L_n=\!\!\!\!\sum_{\substack{n_1,\ldots,n_k \text{ odd}\\ n_1+\ldots n_k=n}}\!\!\! 4^k\cdot s_{n_1,\ldots,n_k}.$$
Preliminary lemma: the number of permutations of $\{1,\ldots,k\}$ such that every element has an odd order is given by: $$ C_k = k!\cdot [x^k]\frac{1}{1-x}\cdot\exp\left(-\frac{x^2}{2}-\frac{x^4}{4}-\ldots\right)= k!\cdot [x^k]\sqrt{\frac{1+x}{1-x}}. $$
Consequence: the number of coloured permutations is given by $$ \sum_{c+m+y+k=n}\binom{n}{c}\binom{n-c}{m}\binom{n-c-m}{y} C_c C_m C_y C_k = n!\cdot [x^n]\left(\frac{1+x}{1-x}\right)^2=4n\cdot n!.$$
Done. Double counting wins again.
Another chance would be given by showing that $L_{n+1}-L_n = (n+1)!-n!$ by some recursive argument. However, I was not able to find it.