We want to show that
\begin{align*} G\left( \frac{\partial}{\partial b}\right) F(b) = F\left( \frac{\partial}{\partial x}\right)\left[G(x) e^{xb} \right]\Big|_{x=0} \tag{1} \end{align*}
By assuming that the functions $F$ and $G$ are expandable in a convergent power series around $0$.
As we are assuming that $F$ and $G$ can be Taylor expanded we have
\begin{align*} G(x) = \sum_{n=0}^{\infty} g_n x^n, \qquad F(x) = \sum_{m=0}^{\infty} f_m x^m \end{align*}
Hence, the LHS of $(1)$ is given by
\begin{equation*} G\left( \frac{\partial}{\partial b} \right) F(b) = \sum_{n}^{\infty} \sum_{m}^{\infty} g_n \left( \frac{\partial}{\partial b}\right)^n f_m b^m \tag{2} \end{equation*}
And the RHS of $(1)$ by
\begin{equation*} F\left( \frac{\partial}{\partial x} \right) \left[ G(x) e^{xb}\right]\Big|_{x=0} = \sum_{n}^{\infty} \sum_{m}^{\infty} f_m \left( \frac{\partial}{\partial x}\right)^m \left[g_n x^n e^{xb}\right]\Big|_{x=0} \tag{3} \end{equation*}
The issue is that I do not see how to show equality between $(2)$ and $(3)$. It looks like I should use the product rule on (3) but I do not see how the result would lead to show the equality.