I think the case when the coset type of $\pi$ is $(n)$ can also be treated in the way you treated coset type $(1^n)$.
Namely, take some $i_1$. Its image under $\pi$, call it $i_2$, can be anything, so $2n$ possibilities.
The image of $\sigma(i_1)$ cannot be $i_2$ nor $\sigma(i_2)$, on account of the coset type, so $(2n-2)$ possibilities.
The next step is similar, just replacing $n\to n-1$.
The orbit of $i_1$ under $\pi$ will then lead to $(2n)!!$ while the orbit of $\sigma(i_1)$ will lead to $(2n-2)!!$.
Total number is $(2n)!!(2n-2)!!=n!2^{n-1}\times(n-1)!2^n$ as you have.