Is there something remarkable about the following map between two kinds of expansion $f(x)=\sum_n f_n x^n$ and $\tilde{f}(x)=\sum_n \frac{f_n x^n}{n!}$ ?

I don't know your motivation, but the passage between the two expansions is compatible with the theory of "$q$integers". Recall that an integer $n$ can be written as the sum $1+1+\cdots+1$ ($n$ times). Its "$q$analog" is the number $[n]_q = 1 + q + \cdots + q^{n1}$. This is $n$ when $q=1$ and $1$ when $q=0$. Similarly, $n!$ has a $q$analog given by $[n]_q! = \[1\]_q \cdot [2]_q \cdots [n]_q = (1) \cdot (1+q) \cdot (1+q+q^2) \cdots (1+q+\cdots +q^n)$. If $q=1$ this is the usual factorial, and if $q=0$ it is simply $1$. Thus if you consider $$f_q(x) = \sum \frac{f_n}{[n]_q!} x^n,$$ then $q=0$ corresponds to your $f$ and $q=1$ to your $\tilde{f}$. There are some surprising extensions of a number of combinatorial objects to $q\neq 1$, see for example the Wikipedia page on qanalogs. For example, if $\tilde{f}$ satisfies a differential equation, then $f$ satisfies an appropriate $q$analog of it (this was effectively mentioned in one of the comments, as the $q$derivative at $q=0$ is simply multiplication by $1/x$). 

