Consider the exponential generating function for a sequence $\mathbf{a}$, given by:
$$\text{EG}_\mathbf{a}(x) = \sum_{n=0}^\infty a_n \frac{x^n}{n!}.$$
In order to find the sequence $\mathbf{a}$ that corresponds with certain specified forms for the generating function, it would be useful to be able to "invert" this expression to a form like this:
$$\frac{x^n}{n!} = \sum_{k \in \mathbb{Z}} b_{n,k} e^{kx} \quad \quad \quad \text{for all }n=0,1,2,....$$
In this latter form we have an infinite matrix $\mathbf{b} = [b_{n,k} ]$ with indices $n =0,1,2,...$ and $k \in \mathbb{Z}$. In this expression, a weighted sum of exponentials gives a single term in the exponential generating function. Taking $c_k \equiv \sum_{n=0}^\infty a_n b_{n,k}$ we can write the generating function $\text{EG}_\mathbf{a}(x) = \sum_{k \in \mathbb{Z}} c_k e^{kx}$, so this form allows us to obtain an alternative expression for the generating function using a weighted sum of exponentials at regular intervals.
My question: Is there any $\mathbf{b}$ that solves with set of equations? If so, how do we find it? (If there is a solution in only some cases, what cases are those?)
My working so far: Suppose we can find a matrix $\mathbf{b}$ which satisfies:
$$\sum_{k \in \mathbb{Z}}^\infty b_{n,k} k^r = \mathbb{I}(r=n) \quad \quad \quad \text{for all } r,n =0,1,2,....$$
Then we have:
$$\begin{equation} \begin{aligned} \sum_{k \in \mathbb{Z}}^\infty b_{n,k} e^{kx} &= \sum_{k \in \mathbb{Z}}^\infty b_{n,k} \sum_{r=0}^\infty \frac{(kx)^r}{r!} \\[6pt] &= \sum_{r=0}^\infty \frac{x^r}{r!} \sum_{k \in \mathbb{Z}}^\infty b_{n,k} k^r \\[6pt] &= \sum_{r=0}^\infty \frac{x^r}{r!} \cdot \mathbb{I}(r=n) \\[6pt] &= \frac{x^n}{n!}, \\[6pt] \end{aligned} \end{equation}$$
which gives the desired form. This means that the desired form can be obtained by solving a series of equations involving sums of powers. I'm not sure if this constitutes "progress" since this set of equations seems just as difficult to solve as the initial problem.
This is not an answer.
Suppose that we have a limited summation $$y=\sum_{i=0}^n a_i \frac {x^i}{i!}$$ that we could consider as a truncated Taylor series $$y=\sum_{i=0}^n a_i \frac {x^i}{i!}+O(x^{n+1})$$ Using series reversion, we could write $$x=\sum_{i=1}^n b_i t^i+O(t^{n+1})\qquad \text{where}\qquad t=\frac{y-a_0}{a_1}$$ and then use the binomial expansion to $$\frac{x^p}{p!}=\frac{1}{p!}\left(\sum_{i=1}^n b_i t^i+O(t^{n+1}) \right)^p$$ in which, later, we should to re-expand the powers of $t$ in terms of $y$ to finally obtain $$\frac{x^p}{p!}=\sum_{i=0}^n c_i y^i+O(y^{n+1})$$
Trying to do it with $y=e^x$ and $n=6$, we should obtain $$\frac {x^1}{1!}=-\frac{49}{20}+6 y-\frac{15 y^2}{2}+\frac{20 y^3}{3}-\frac{15 y^4}{4}+\frac{6 y^5}{5}-\frac{y^6}{6}+O\left(y^7\right)$$ $$\frac {x^2}{2!}=\frac{469}{180}-\frac{223 y}{20}+\frac{879 y^2}{40}-\frac{949 y^3}{36}+\frac{41 y^4}{2}-\frac{201 y^5}{20}+\frac{1019 y^6}{360}+O\left(y^7\right)$$ $$\frac {x^3}{3!}=-\frac{267}{160}+\frac{349 y}{36}-\frac{18353 y^2}{720}+\frac{797 y^3}{20}-\frac{1457 y^4}{36}+\frac{4891 y^5}{180}-\frac{187 y^6}{16}+O\left(y^7\right)$$ $$\frac {x^4}{4!}=\frac{95}{128}-\frac{7667 y}{1440}+\frac{24901 y^2}{1440}-\frac{4013 y^3}{120}+\frac{122249 y^4}{2880}-\frac{5273 y^5}{144}+\frac{10279 y^6}{480}+O\left(y^7\right)$$ $$\frac {x^5}{5!}=-\frac{8591}{34560}+\frac{6041 y}{2880}-\frac{92771 y^2}{11520}+\frac{13349 y^3}{720}-\frac{163313 y^4}{5760}+\frac{43319 y^5}{1440}-\frac{129067 y^6}{5760}+O\left(y^7\right)$$
May be, this could give you some ideas (I hope and wish).