Let $\mathcal{A}$ be a nonempty set of pairwise coprime positive integers, and for each $a \in \mathcal{A}$ let $\Omega_a \subsetneq \{0, \ldots, a - 1\}$ be a set of residues modulo $a$. Furthermore, put $$\mathcal{S} := \{n \in \mathbf{N} : (n \bmod a) \notin \Omega_a, \;\forall a \in \mathcal{A}\}.$$ An easy application of the Chinese Remainder Theorem shows that $$\underline{\mathbf{d}}(\mathcal{S}) \geq \prod_{a \in \mathcal{A}} \left(1 - \frac{\#\Omega_a}{a}\right),$$ where $\underline{\mathbf{d}}$ denotes the lower asymptotic density. In particular, if $$\sum_{a \in \mathcal{A}} \frac{\#\Omega_a}{a} < +\infty$$ then the infinite product converges and $\mathcal{S}$ has positive asymptotic density. (Actually, the asymptotic density of $\mathcal{S}$ exists and it is equal to the infinite product, but this is irrelevant for my question.)

My question is: What if we drop the hypothesis that the elements of $\mathcal{A}$ are pairwise coprime? Are there results that guarantee that $\mathcal{S}$ has a positive lower asymptotic density? Have these kind of sieves been studied?

Of course, if the elements of $\mathcal{A}$ are not pairwise coprime then (without additional hypotheses) $\underline{\mathbf{d}}(\mathcal{S})$ can be $0$, just consider: $\mathcal{A} = \{2, 4\}$, $\Omega_2 = \{0\}$, $\Omega_4 = \{1, 3\}$.

Thank you for any suggestion.