Absolutely continuous coupling of probability measures

Let me formulate and prove it in greater generality (which actually makes your question easier). Let $X$ be a metric space, and $\mu$ be a probability measure on $X\times X$ (for simplicity I consider the product of two copies of $X$ only; the general case is precisely the same). You want to obtain a sequence of measures $\nu_n$ on $X\times X$ which

(1) have the same marginals $\mu_1,\mu_2$ as $\mu$;

(2) are absolutely continuous with respect to the product measure $\mu_1\times\mu_2$;

(3) weakly converge to $\mu$.

For each $n$ take a countable partition of $X$ into measurable sets $X^i$ with diameter $\le 1/n$ (presuming the space $X$ is such that partitions like this exist for any $n$) and denote by $\mu_\epsilon^i$ the normalized restriction of the measure $\mu_\epsilon$ to $X^i$. Then put $$ \nu_n = \sum_{i,j} \mu(X^i\times X^j) \mu_1^i\times \mu_2^j \;. $$

There are various papers where this question occurs. I guess a paper which directly covers the case you are interested in is https://arxiv.org/pdf/1901.07407.pdf . Note that here, the marginals don't have to be one dimensional.

A more general case in polish spaces is studied in https://arxiv.org/pdf/1811.00304.pdf Proposition 2. Since there is no Lebesgue measure on polish spaces, here the question is about absolute continuity with respect to the product measure of the marginals. Since in your case the product measure is absolutely continuous with respect to the Lebesgue measure, this of course also answers your problem.

There are other papers which incorporate this problem, mostly in the context of regularized optimal transport, but I do not have the references at hand.

Note also that in your case, with one dimensional marginals, things are simple as the concept of quantile functions and copulas can be used (similar in spirit to Iosif's answer) .