There is a number of equivalent ways to define a distributive law between monads $(S, ν, ϑ)$ and $(T, μ, η)$. From the elementary defintion as a compatible transformation $λ : ST → TS$, to slicker defintions as a monad in the category of monads, or liftings of $T$ to a monad on $X^S$.
Another approach is to start from a monad structure $(TS, ψ, ζ)$ on $TS$ and ask when does it come form a distributive law of $S$ over $T$, or more concretely, when is $λ := ψ ∘ ηSTϑ$ that distributive law. Of course we don't expect the answer to be always (and it isn't), but then the question of necessary conditions arises.
Accodring to this post, there is a very simple and satisfying answer: it is enough for $ηϑ$ to be the unit of $TS$, $ψ ∘ TϑηS = TS$, and $ψ ∘ TϑTϑ = μϑ$ and $ψ ∘ ηSηS = ην$ (ie. $Tϑ$ and $ηS$ are monoid morphisms, so that $ψ$ "restricts" to $μ$ and $ν$ respectively). If this is so, then we in particular have that the monad multiplication $μν ∘ TλS = μν ∘ TψS ∘ TηSTϑS$ defined via $λ$ equals the original multiplication $ψ$. Unfortunately, I'm having trouble proving this. Obviously you need to expand some of the factors in $TηSTϑS$ further via $ψ$ so that you can use associativity to rearange them into something you can work with, but I don't see how.
There is an analogous result on page 261 of Toposes, Triples and Theories, but the assumptions are stronger. (C1)-(C3) are stronger versions of those in the previous paragraph, and the additional (C4) and (C5) let you freely commute $ψ$ with $ν$ and $μ$, from which we immediately have the required $μν ∘ TψS ∘ TηSTϑS = ψ ∘ μSTν ∘ TηSTϑS = ψ$.
Well, it turns out that I should have thought about this a bit more before asking the question. In case anyone else is interested, here's the proof that if $Tϑ$ and $ηS$ are monoid morphisms and $ϑη$ a "middle" unit as described in the question, then the stronger conditions of Toposes, Triples and Theories also follow, and from there it's easy to show that $ψ$ comes from a distributive law.
First, we can strengthen the condition $ψ ∘ ηSηS = ην$ to $ψ ∘ TS ηS = Tν$. By inserting units we have $ψ ∘ TSηS = ψ_4 ∘ Tϑ ηS ηϑ ηS$, where $ψ_n$ is the canonical morphism $(TS)^n → TS$, and from there $$= ψ_3 ∘ TS ψ TS ∘ Tϑ ηS ηϑ ηS = ψ_3 ∘ Tϑ ηS ηS = ψ ∘ Tϑ ηS ∘ Tν = Tν.$$
This and the symmetric argument for $ϑ$ are the conditions (C2) and (C3). To get (C4) and (C5), ie. $ψ ∘ TSTν = Tν ∘ ψS$ and analogously for $μ$ we can now just use $Tν = ψ ∘ TS ηS$, associativity of $ψ$, and $ψ ∘ TS ηS = Tν$ again.
As a side note, a useful toy example that was for me very helpful in sorting this out are monoids in $\mathrm{Set}$, where distributive laws $l: M × N → N × M$ correspond to Zappa-Szép products. In particular, in case of groups the equivalence boils down to a well known fact: if $K, H ≤ G$, $K ∩ H = 1$ and $KH = G$, then $G$ is isomorphic via the canonical isomorphism $UG ≅ UK × UH$ to a Zappa-Szép product.