Actually, you can still derive counting formulas for unambiguous regular expressions with Kleene stars within.
Given the inductive definition of a regular expression as:
$$
\begin{equation*}
e \in \mathrm{Re} := x \in \Sigma \mid e_0 ~ e_1 \mid e_0 + e_1 \mid e^*
\end{equation*}
$$
Consider the following translation $[\![\cdot]\!] : \mathrm{Re} \to \mathbb{C}(z)$ that takes a regular expression and translates it into a complex-valued rational function:
$$
\begin{align*}
[\![x \in \Sigma]\!] &= z \\
[\![e_0 ~ e_1]\!] &= [\![e_0]\!] \times [\![e_1]\!]\\
[\![e_0 + e_1]\!] &= [\![e_0]\!] + [\![e_1]\!]\\
[\![e^*]\!] &= \frac{1}{1 - [\![e]\!]}
\end{align*}
$$
We can show that this translation returns a rational expression by doing structural induction on $e$, and noting that all of the operations used on the right-hand side preserves rational-ness.
Suppose that the regular expression $e$ that we put in is unambiguous, then we would find that the rational function denoted by $[\![e]\!] \in \mathbb{C}(z)$ is actually the generating function for the family of words that are accepted by the language underlying $e$, ranked by their length.
For example, consider the language $(a^*b)^*$, which defines the language of runs of $a$ delimited by $b$. Now, this regular expression is unambiguous, so we can run our translation trick:
$$
\begin{align*}
[\![(a^*b)^*]\!] &= \frac{1}{1 - [\![a^*b]\!]} \\
&= \frac{1}{1 - ([\![a^*]\!] \times [\![b]\!])} \\
&= \frac{1}{1 - \left(\frac{1}{1 - [[a]]} \times z\right)} \\
&= \frac{1}{1 - \frac{z}{1 - z}} \\
&= \frac{1}{2} + \frac{1}{2 - 4 z}
\end{align*}
$$
As it turns out, given the above generating function, its coefficient extraction will be
$$
[z^n][\![(a^*b)^*]\!] = 2^{n - 1} + \frac{\delta\left(n\right)}{2}
$$
where
$$
\delta(n) = \begin{cases}
1 & \text{if } n = 0 \\
0 & \text{otherwise}
\end{cases}
$$
In fact, since our translation $[\![\cdot]\!]$ generates rational functions, we can use a partial fraction decomposition to create an enumeration formula for any unambiguous regular expression.
Suppose you have a irreducible rational function
$$
r(z) + \frac{p(z)}{q(z)}
$$
where $r, p, q$ are polynomials, then you can decompose this into
$$
r(z) + \frac{C_0}{z - q^*_0} + \dots + \frac{C_n}{z - q^*_n}
$$
where $q^*_k$ are the roots of $q(z)$. There's a bit of technical corner-cases (like multiplicity of roots, etc), but it's relatively easy to do coefficient extraction on the expression above:
$$
[z^n] \frac{C}{z - q^*} = C \times {q^*}^{-n}
$$
In fact, the partial fraction decomposition generalize to multivariate rational functions, so you can actually construct counting formulas for queries such as "How many words are there where there are $n$ as and $m$ bs?"
Unfortunately, the extent to which this method will be useful ends when you have an ambiguous expression.
