Has the following operation $*$ on infinitely differentiable functions $f,g$ (without pole at $x=0$) been studied before?

$$(f*g)^{(n)}(0) = f^{(n)}(0) \cdot g^{(n)}(0)$$

where $n$ is a nonnegative integer? This is equivalent to defining

$$(f*g)(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0) \cdot g^{(n)}(0)}{n!}x^n.$$

If $f$ and $g$ are treated as generating functions, then $(f*g)^{(n)} = (f^{(n)} \otimes g^{(n)}) \cdot n! $, where $\otimes$ is the typical Hadamard product for series.

Clearly the identity is $\exp$, our product is associative and commutative — this means infinitely differentiable functions without pole at $x=0$ form a monoid under our product.

Other straightforward identities (forgive my abuse of notation):

• $e^{ax} * e^{bx} = e^{abx}$;
• $(cf)*g = f*(cg) = c(f*g)$;
• $\cosh x * e^{ix} = \cos x$; and
• If $a_n = \frac{f^{(n+1)}(0)}{f^{(n)}(0)}$, then the exponential generating function $\mathrm{EG}(a_n; - ) * f = f'$.