I would like to rewrite the series $$\sum_{n=0}^\infty \frac{1}{n!}(\Delta^\varepsilon)^n a_n,$$ where $\Delta^\varepsilon=\sum_{k=1}^\infty \varepsilon^k b_k$, as a series in $\varepsilon$ $$\sum_{n=0}^\infty c_n \varepsilon^n$$ (i.e., I need to find the $c_n$).
Obviously, I can do this term-by-term. But the general case seems quite difficult.
We have, using Cauchy products, at least for $\varepsilon$ small enough, that
$$(\Delta^{\varepsilon})^n=\sum_{k=1}^{+\infty}\left(\sum_{j_1+\dots+j_n=k}\prod_{i=1}^nb_{j_i}\right)\varepsilon^k,$$
hence, if the coefficients have a good behaviour
\begin{align}
\sum_{n=0}^{+\infty}\frac 1{n!}(\Delta^{\varepsilon})^na_n&=\sum_{n=0}^{+\infty}\frac 1{n!}a_n\sum_{k=1}^{+\infty}\left(\sum_{j_1+\dots+j_n=k}\prod_{i=1}^nb_{j_i}\right)\varepsilon^k\\\
&=\sum_{k=1}^{+\infty}\sum_{n=0}^{+\infty}\frac 1{n!}a_n\left(\sum_{j_1+\dots+j_n=k}\prod_{i=1}^nb_{j_i}\right)\varepsilon^k.\\\
\end{align}
So we can take
$$c_k:=\sum_{n=0}^{+\infty}\frac 1{n!}a_n\left(\sum_{j_1+\dots+j_n=k}\prod_{i=1}^nb_{j_i}\right)$$
for $k\geq 1$ and $c_0=0$.
