Ok, I think I've solved the mystery, and it is a little disappointing: The point is that there is actually a distributive law lurking in the background , constructed from the $l$ in the original question, that gives the same monad structure on the composite $ST$. (Many thanks to მამუკა ჯიბლაძე and Varkor - I've realized it by trying to follow their surgestions in the comment.)


Right, so the idea is that given a $l : TS \to ST$ as in the original post, I consider the following map $\alpha: TS \to ST $ defined as the composite:

$$ TS \overset{\epsilon_S TS }\to STS \overset{l^{-1}S}\to TSS \overset{T \mu_S}\to TS \overset{l}\to ST $$

Then I claim that this $\alpha$ is a distributive law and that the monad structure on $ST$ defined from $l$ in the original post can be obtained from $\alpha$ by using the classical formula. The proof is just a bunch of diagram computation that would be hard to reproduce here (and it is a little late so I hope I didn't make any mistakes in my computation)

For example, in the case of $M$ and $L(M)$ in the original question, $\alpha$ is the map $L(M) \times M \to M \times L(M)$ defined by $\alpha((m_1,\dots,m_k),m)  = (1 , (m_1,\dots,m_k m))$


It should be noted that while $l$ was invertible, $\alpha$ no longer is. So there seem to be still something interesting going on here. For example, I have the impression that one can also define a distributive law $ ST \to TS$ using something similar that produce the monad structure on $TS$ coming from the isomorphism $ST \simeq TS$ given by $l$ and the monad structure on $ST$). But in any case that solve my problem: This is "just" a distributive law with some special additional property, but I was using the wrong function $TS \to ST$.