Is there a combinatorial proof that the number of fair permutations of $\{1,2,\dots,n\}$, with exponential generating function $e^x\sec x$, is divisible by $2^{\lfloor n/2\rfloor}$?

An interesting fact about these identities is that the coefficients in the expansion of $kf_n$ are the same as the coefficients in the expansion of $k$ as a sum of nonconsecutive powers of the golden ratio $\phi=(1+\sqrt5)/2$. Thus \begin{align*} 1&=\phi^0\\ 2&=\phi^{-2}+\phi\\ 3&=\phi^{-2}+\phi^2\\ 4&=\phi^{-2}+\phi^0+\phi^2 \end{align*} and so on.

The sum is equal to $$ \frac{1}{s}\left(\frac{x}{1+\sqrt{1-x^2}}\right)^s. $$ I don't think there is any connection with Bessel functions.

The right side is equal to $4\,{}_3F_2(1/2,1/2,1/2;1,1\mid 1)$, which can be evaluated by Dixon's theorem.

@TomCopeland I have only one paper on umbral calculus. My approach could be described as applications of linear functionals on polynomials to identities for generating functions. I have not made any connections to geometry or topology, and I haven't done anything with Sheffer sequences.

@TomCopeland Yes, you did demonstrate this explicitly in your answer.

If you write $s(u,z)$ as $\exp(u(\log (1+z))$ then the inverse relation becomes clear—$\log(1+z)$ is the compositional inverse of $e^z-1$. More generally, if $f(z)$ and $g(z)$ are compositional inverses then we have similar inverse matrices formed from the coefficients of $\exp(u f(z))$ and $\exp(u g(z))$.

Maple does produce a solution with the boundary conditions, but it is singular at 0.