In his thesis Ira Gessel provides a method which gives a combinatorial interpretation of the inverse of a counting series of a linked set. A counting series is a generalization of a generating function. Some applications on the technique in the thesis are to things like symmetric functions and formal power series in countably many variables. For the single variable case, see Theorem 4.5 on page 45 which computes the inverse of a single variable generating function for some lattice paths. I will breifly summarize and give some simple examples below to make this answer a little more self-contained.
Let $P$ be a set and $S \subseteq P^*$ be a subset of the free monoid (we actually want $S$ be a linear system which is defined in the thesis). For any $V \subseteq S$ we define its conuting series to be $\Gamma(V) = \sum_{\alpha \in V} \alpha$ which is a formal sum. We also define the alternating counting series $\bar{\Gamma}(V) = \sum_{\alpha \in V} (-1)^{r(\alpha)} \alpha$ where $r(\alpha)$ is the number of prime factors in $\alpha$ (i.e. length of the word). For any $\alpha = a_1a_2 \cdots a_n$ the links of $\alpha$ are $a_i a_{i+1}$. Take some set a links $L \subseteq P^2$ and define $C$ to be the set of $\alpha \in S$ such that all links of $\alpha$ are in $L$. Also let $\bar{C}$ be the set of $\alpha \in S$ such that all links of $\alpha$ are not in $L$. On page 37 we have Theorem 4.1 (The Inversion Theorem) we states $\Gamma(C) \bar{\Gamma}(\bar{C}) = 1$. Now let's use this for some single variable generating functions.
For a trivial example let $S = P^*$ where $P$ is a set of size $k$. Then let $L = P^2$. In this case $\Gamma(C) = \sum_{w \in P^*} w$ is the sum of all $k$-ary words and $\bar{\Gamma}(\bar{C}) = 1 - \sum_{a \in P} a$. If we subsitute $a = x$ for all $a \in P$ we get the $\Gamma(C)$ becomes $\sum_{n \geq 0} k^n x^n$ and $\bar{\Gamma}({\bar{C}})$ becomes $1 - kx$.
For a less trivial example let $S = P^* = \{a,b\}^*$ and let $L = \{aa,ab,ba\}$. Then $\Gamma(C)$ is the sum of all binary words which have no consecutive $b$'s and $\bar{\Gamma}(\bar{C}) = 1 - a - b + b^2 - b^3 + \cdots$. Again substituting $a = b = x$ we have
$$1 + 2x + 3x^2 + 5x^3 + \cdots = \frac{1}{1 - 2x + x^2 - x^3 + \cdots}$$
the inverse of an "almost" Fibonacci generating function.