The question is very interesting, and Chris Jones gave a very nice answer (not ugly at all!) that can be interpreted in terms of partitions, as follows.
Without loss of generality, we can assume the zeroth coefficient to be $a_0=1$, so that we are inverting:
$$F = 1 + \displaystyle\sum_{n \geq 1} a_nx^n ,$$
and then
\begin{align*}
F^{-1} &= \displaystyle\sum_{n \geq 0} (-1)^n \Big(\sum_{i \geq 1} a_{i}x^i\Big)^n
\end{align*}
(a consequence of the geometric series expansion: $\frac1{1+w} = \sum_{n\geq 0} (-1)^n w^n$ with $w=\sum_{i \geq 1} a_{i}x^i$.)
Next, use the multinomial theorem to expand the inner sum:
\begin{align*}
F^{-1} &= \displaystyle\sum_{n \geq 0} (-1)^n \Big(\sum_{i \geq 1} a_{i}x^i\Big)^n\\
&= \displaystyle\sum_{n \geq 0} (-1)^n \Big(\sum_{\substack{ \beta_1, \beta_2,\dots,\\\sum_{i} \beta_i = n}} \binom{n}{\beta_1, \beta_2,\dots} \prod_{i \geq 1} (a_{i}x^i)^{\beta_i} \Big)\\
&= \displaystyle\sum_{n \geq 0}\sum_{\substack{\beta_1, \beta_2,\dots,\\\sum_{i} \beta_i = n}} \Big((-1)^n\binom{n}{\beta_1,\beta_2,\dots}\prod_{i \geq 1} a_{i}^{\beta_i}\Big) x^{\sum_i i\beta_i}
\end{align*}
where the inner sum is over all natural number sequences $\langle \beta_i\rangle$, and $\binom{n}{\beta_1,\beta_2, \dots}$ is a multinomial coefficient. Grouping terms by exponent on $x$, we have the somewhat-closed form
$$F^{-1} = \sum_{n\geq 0} b_n x^n = \displaystyle\sum_{n \geq 0} \Bigg(\sum_{\substack{\beta_1, \beta_2, \dots\\\sum_{i}i\beta_i= n}} (-1)^{\sum_i \beta_i}\binom{\sum_i \beta_i}{\beta_1, \beta_2, \dots} \prod_{i \geq 1} a_i^{\beta_i}\Bigg) x^n $$
Finally, the formula for $F^{-1}$ can be rewritten in terms of partitions of the natural number $n$.
Indeed, partitions of $n$ (into equal or different non-zero parts) are in bijection with sequences of non-negative integers $(\beta_1,\cdots,\beta_n)$, satisfying $$\sum_{i}i\beta_i= n.$$
This gives the interpretation of each $\beta_i$ as the number of parts of size equal to $i$ (it can, of course, be zero). Moreover $\sum_{i}\beta_i$ becomes now the total number of parts of the partition (usually called the length of the partition).
Then, the formula for the coefficients $b_n$ of $F^{-1}$ is:
$$b_n = \sum_{\beta=(\beta_i)\in P_n} (-1)^{|\beta|} |\beta| ! \prod_{i=1}^n \frac{a_i^{\beta_i}}{\beta_i !} $$
where $P_n$ denotes the finite set of all partitions of $n$, and $|\beta|$ is the length of the partition $\beta = (\beta_1,\cdots,\beta_n)\in P_n$
