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 $TS \to ST$.


**The case of monoids:** Following the suggestion of მამუკა ჯიბლაძე in the comment, here is what happens with distributive law between ordinary monoids, which I think gives a good idea of what happens in general:

In the standard case, a distributivity law of $T$ on $S$ can be thought of as a monoid $M$ that contains $S$ and $T$ as submonoids and such that each element of $M$ is written uniquely as $st$. The distributive law $TS \to ST$  itself is the function that gives you the expression as $st$ of an element of the form $ts$. You then use it to describe the multiplication $(s_1 t_1)(s_2 t_2) = (s_1 s'_2) (t'_1 t_1)$

The case described in the original post corresponds to the situation where we have two inclusions of monoids $i,j: T \to M$, and $S$ is a submonoid of $M$, and every element of $M$ can be written uniquely both as $j(t) s$ and as $s' i(t')$. The function $l$ is now the bijection that sends $(t,s)$ to $(s',t')$ where $j(t) s =  s' i(t')$.

One can also use it to compute the product a bit like a distributivity law, but one has to be a bit more careful on how we do it:

$$(s_1 i(t_1)) (s_2 i(t_2)) = j(t'_1) s_1 s_2 i (t_2) =s'' i(t''_1) i(t_2) = s'' i(t''_1 t_2) $$ 

which corresponds exactly to the formula for the product in the original question.

The fact that this $l$ isn't compatible to unit law of $S$ exactly means that $i \neq j$, indeed $i$ and $j$ are respectively obtained as $T \overset{\epsilon_s T}\to ST$ and $T \overset{T \epsilon_s }\to TS \overset{l}{\to} ST$ so the compatibility of $l$ with $\epsilon_S$ is exactly the condition that $i=j$. The compatibility of $l$ with multiplication fails for similar reason: if one try to swap $j(t)$ with $s_1 s_2$ one can do it in a single step, but doing in two steps doesn't make sense as $j(t)s_1 s_2 = s'_1 i(t') s_2$ and then we are stuck and using $l$ to replace $(t',s_2)$ by a $(s'_2,t'')$ won't give the correct result. 


But nonetheless, in this situation any "wrong side" product $i(t) s$ can be rewritten in the correct order as $s'i(t')$ by computing the product $(1 i(t)) (s i(1))$ using the formula given above, and this is how one obtains the distributive law $\alpha$ mentioned above.