Given $f \in C^{\infty}(\mathbb{C})$ we can assign to it a power series (which may or may not converge to $f$) $\sum_{0}^{+\infty}a_n z^n$. Now, Taylor's theorem tells us that if a function can be developed as a power series then the term $a_n$ must be of the form $\frac{f^{(n)}(0)}{n!}$.
Wanting to investigate a continuous version of Tayor's series one might consider this intuitive modification $$\sum_{0}^{+\infty}a_n z^n \longrightarrow \int_{0}^{+\infty}a(t) z^t dt$$ After manipulating slightly the exponential we end up with this expression $$\int_{0}^{+\infty}a(t) e^{-st} dt \mbox{ , } s \in \mathbb{C} \mbox{, } \mbox{Re}(s) \in \mathbb{R}^+ \mbox{, } \mbox{Im}(s) \in \mathbb{R}$$ All explanations that I found usually end with something like "and this is how we define the Laplace trasform of the function $a(t)$". The function inside of the integral does not have a specific form, it is simply a signal that we can choose to insert into the expression to compute its Laplace transform.
So my question ultimately is, could something like Taylor's theorem exist in the continuous case? Fractional derivatives exist and factorials can be extended to real numbers, so why can't something like a "Taylor integral" or "Taylor transform" be defined? $$f(z)=\int_0^{+\infty}a(t) z^t dt$$ And just like with Taylor series we would say that if $f$ can be represented as a "Taylor integral" then $a(t) = \frac{f^{(t)}(0)}{\Gamma(t+1)}$, giving: $$ f(z)=\int_0^{+\infty}\frac{f^{(t)}(0)}{\Gamma(t+1)} z^t dt \mbox{, or alternatively } f(s)=\int_0^{+\infty}\frac{f^{(t)}(0)}{\Gamma(t+1)} e^{-st}dt$$