Yes, here's one way to do it. Suppose $A(x) = \frac{P(x)}{Q(x)}$ and $B(x) = \frac{R(x)}{S(x)}$ where $P, Q, R, S$ are polynomials, and WLOG $Q$ and $S$ have constant term $1$. Write
$$Q(x) = \prod_i (1 - \lambda_i x), S(x) = \prod_j (1 - \mu_j x).$$
Then we can write $a_n$ as a sum of terms of the form $p(n) \lambda_i^n$ where $p$ is a polynomial, and similarly for $b_n$ and $\mu_j$. The product of two such terms is another such term, but with an exponential with base $\lambda_i \mu_j$. From this you can show that $(A \star B)(x)$ is rational with denominator
$$(Q \otimes S)(x) = \prod_{i, j} (1 - \lambda_i \mu_j x).$$
This polynomial can be computed fairly explicitly as
$$\det (I - (M \otimes N) x)$$
where $\otimes$ is the Kronecker product or tensor product of matrices and $M$ is the companion matrix of $x^{\deg Q} Q \left( \frac{1}{x} \right)$, and similarly for $N$ and $S$; in particular its coefficients are a polynomial in the coefficients of $Q$ and $S$. Once you know the denominator, the numerator is determined by compatibility with initial conditions.
Edit, 6/24/21: I've sketched this argument out in a bit more detail here.
For example, suppose we wanted to compute the Hadamard square of the generating function $\frac{x}{1 - x - x^2}$ of the Fibonacci numbers, namely
$$\sum_n F_n^2 x^n = x + x^2 + 4x^3 + 9x^4 + 25x^5 + \dots$$
The relevant companion matrix is $M = \left[ \begin{array}{cc} 0 & 1 \\ 1 & 1 \end{array} \right]$, and its tensor square is
$$M \otimes M = \left[ \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 \\ 1 & 1 & 1 & 1 \end{array} \right].$$
We compute that
$$\det (I - (M \otimes M) x) = 1 - x - 4x^2 - x^3 + x^4$$
so this is the denominator of the generating function. We can compute the numerator by multiplying the denominator by the first few terms of the series $\sum F_n^2 x^n$ which gives
$$(1 - x - 4x^2 - x^3 + x^4)(x + x^2 + 4x^3 + 9x^4 + \dots) = x - x^3$$
which gives the final answer
$$\boxed{ \sum F_n^2 x^n = \frac{x - x^3}{1 - x - 4x^2 - x^3 + x^4} }.$$
You can check this to however many terms you like in WolframAlpha.
You can also bypass the last step of computing the numerator using initial conditions by doing a little extra work ahead of time. First, show that every series with a rational generating function (edit: whose numerator has degree less than its denominator) can be written in the form
$$a_n = u^T M^n v$$
for some square matrix $M$ and some vectors $u, v$ (you can take $M$ to be the companion matrix above which is why I'm using the same letter for it). The generating function of such a series can therefore be written
$$A(x) = u_a^T (I - Mx)^{-1} v_a$$
which you can compute explicitly using Cramer's rule, which among other things shows that the denominator of $A(x)$ is $\det (I - Mx)$ as expected. Suppose we've also written $b$ this way, so
$$b_n = u_b^T N^n v_b, B(x) = u_b^T (I - Nx)^{-1} v_b.$$
Then it's an exercise to show that
$$a_n b_n = (u_a \otimes u_b)^T (M \otimes N)^n (v_a \otimes v_b)$$
from which it follows that
$$\boxed{ (A \star B)(x) = (u_a \otimes u_b)^T (I - (M \otimes N) x)^{-1} (v_a \otimes v_b) }.$$
This is perhaps the most conceptual proof that the Hadamard product of two rational power series is rational, and depending on how you've written down $A$ and $B$ it is sometimes also the fastest way to compute it. For example, the Fibonacci numbers are easy to write in this form, in the sense that the vectors $u, v$ are very simple, so the previous calculation can be redone this way without much difficulty.