The Lagrange Reversion theorem states
$$x=a+bf(x)\implies g(x)=g(a)+\sum_{n=1}^\infty\frac{b^n}{n!}\frac{d^{n-1}}{da^{n-1}}(f^n(a)g’(a))$$
Our goal is using the Lagrange reversion formula obtaining:
$$\sum_{n=1}^\infty a_n (n+u)^{vn}b^n,v<0\tag1$$
where $a_n$ does not cancel the $(n+u)^{vn}$. We must have: $$\frac{d^{n-1}}{da^{n-1}}(f^n(a)g’(a))=h(a,n)(n+u)^{vn}\tag2$$ where $h(a,n)$ does not cancel the $(n+u)^{vn}$. Any special case, especially $u=0,v=-1$, will satisfy the question.
One idea:
uses $$\frac{d^{n-1}}{da^{n-1}}\left(\left(e^{a^{-r}}\right)^n\right)=\frac{d^{n-1}}{da^{n-1}}e^{\left(\frac a{\sqrt[r]n}\right)^{-r}}=n^\frac{1-n}r\left.\left(\frac{d^{n-1}}{da^{n-1}}e^{a^{-r}}\right)\right|_{a\to\frac a{\sqrt[r]n}}$$
a series appears:
$$n^\frac{1-n}r\left.\left(\frac{d^{n-1}}{da^{n-1}}e^{a^{-r}}\right)\right|_{a\to\frac a{\sqrt[r]n}}= n^\frac{1-n}r\left.\left(\frac{d^{n-1}}{da^{n-1}}\sum_{k=0}^\infty \frac{a^{-rk}}{k!}\right)\right|_{a\to\frac a{\sqrt[r]n}}= \left.n^\frac{1-n}r\sum_{k=0}^\infty \frac{a^{1-n-kr}(-kr)!}{(1-n-kr)!k!}\right|_{a\to\frac a{\sqrt[r]n}}=\sum_{k=0}^\infty\frac{(-kr)!n^k }{(1-n-kr)!a^{kr+n-1}}$$ where the substitution cancels $n^\frac{1-n}r$ failing the goal. $$\vphantom\ $$ What functions $f(a),g(a)$ solve $(1),(2)$?