This is not an answer, rather a possible suggestion on how to deal with irreducible polynomials $A(x)$: Let $A(x)\in\mathbb F_2[x]$ be irreducible of degree $n$. Then \begin{equation} A(x)=\prod_{i=0}^{n-1}(1+\lambda^{2^i}x) \end{equation} for some $\lambda\in\mathbb F_{2^n}$. The partial fraction decomposition and geometric series yield \begin{equation} \frac{1}{A(x)}=\sum_{i=0}^{n-1}\frac{\alpha^{2^i}}{(1+\lambda^{2^i}x)} = \sum_{m=0}^\infty\sum_{i=0}^{n-1}\alpha^{2^i}(\lambda^{2^i}x)^m = \sum_{m=0}^\infty T(\alpha\lambda^m)x^m, \end{equation} where $\alpha=\lambda/f'(1/\lambda)$ and $T$ is the trace map from $\mathbb F_{2^n}$ to $\mathbb F_2$.
Note that the power series is periodic with period $e$, where $e$ is the multiplicative order of $\lambda$. Thus if $U$ is the subgroup of order $e$ of $\mathbb F_{2^n}^\star$, then the density of $1$'s is the number of $u\in U$ with $T(\alpha u)=1$ divided by $\lvert U\rvert$.
An easy case is when $e=2^n-1$, so $U=\mathbb F_{2^n}^\star$. Half of the elements of $\mathbb F_{2^n}$ have trace $0$, so the density of $1$'s is $2^{n-1}/(2^n-1)$.
So when not only $n$ is prime, but even $2^n-1$ is prime, then we have this case and the density is only slightly bigger than $1/2$.
The general case seems to be more challenging. It is always difficult to relate an additive function like the trace map with subgroups of the multiplicative group of fields.