Showing that exponential of a derivation is automorphism on formal power series ring
You are right that the inverse of $\exp TD$ is $\exp (-T)D$, which sends $g$ to $\sum_{p \in \Bbb N_0} \frac 1{p!} D^p(g)(-T)^p$.
To conclude your proof, you only need the following identity: $$\sum_{p + q = n} \frac 1{p!q!}(-1)^q = \begin{cases}1, &\textrm{if }n = 0;\\0, &\textrm{if }n > 0.\end{cases}$$ This identity becomes obvious when you multiply it by $n!$, which then becomes $\sum_{p + q = n}\binom n q (-1)^q = (1 - 1)^n$.