From: https://en.wikipedia.org/wiki/Generating_function_transformation#OGF_⟷_EGF_conversion_formulas,
It states that if you have a an ordinary generating function of the form $$\displaystyle{\sum_{n=0}^{\infty}} f_n z^n$$, a re-summation technique to evaluate it in terms of its exponential generating function could be done as follows:
$$\displaystyle{\sum_{n=0}^{\infty}} f_n z^n = \displaystyle{\sum_{n=0}^{\infty}} f_n z^n \frac{n!}{n!} = \displaystyle{\sum_{n=0}^{\infty}} \frac{f_n z^n}{n!} \displaystyle{\int_{0}^{\infty}} e^{-x} x^n \, dx= \displaystyle{\int_{0}^{\infty}} e^{-x}\displaystyle{\sum_{n=0}^{\infty}} \frac{f_n (xz)^n}{n!} \, dx = \displaystyle{\int_{0}^{\infty}} e^{-x} \widehat{F}(xz) \, dx$$
where $\widehat{F}(x)$, is the exponential generating function, $ \displaystyle{\sum_{n=0}^{\infty}} \frac{f_n x^n}{n!}$.
What I'm not sure about is when it could be said that one form of summation is equal to another.
That is, if $$\displaystyle{\int_{0}^{\infty}} e^{-x} \widehat{F}(xz) \, dx$$ converges, then what conditions or assumptions would have to be known/made so that we could conclude that $$\displaystyle{\sum_{n=0}^{\infty}} f_n z^n$$ converges to the same value. I'm assuming that it doesn't simply suffice to show that the integral representation converges, as the ordinary generating function could be derived from that integral representation by repeatedly integrating by parts, which could lead to a divergent asymptotic series.