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.
