Let $\mathfrak{S}_n$ be the permutation group on $[n]$. Denote the cardinality of $\{\pi\in\mathfrak{S}_n: \pi^2=id\}$, the set of involutions, by $I(n)$. It is well-known that these numbers have the exponential generating function $$\sum_{n\geq0}I(n)\frac{x^n}{n!}=e^{x+\frac12x^2}.$$
After these preparations, we wish to tackle the problem of computing arbitrary order derivatives of the function $f(x)=e^{x+\frac12x^2}$, which sure get complicated very quickly.
Question. Instead, is it true we could bypass this with $$\frac{D^mf(x)}{f(x)}=(x+I)^m?$$
In umbral notation, one writes $I^k=I(k)$.
I wish to see combinatorial arguments; if not, other cute proofs will do.