In category theory a monad consists in an endofunctor $M\colon \mathcal{C}\rightarrow \mathcal{C}$ together with natural transformations $\mu \colon M \circ M \Rightarrow M$ and $\eta \colon Id_{\mathcal{C}}\Rightarrow M$ satisfying the usual associativity and unitality conditions.
The idea is to generalize the usual notion of monoid to a category not having a monoidal structure: indeed, if $(\mathcal{C}, \otimes, 1)$ is a monoidal category and $A$ is a classical monoid then $A\otimes - \colon \mathcal{C}\rightarrow \mathcal{C}$ is a monad.
For a monad $M$ we can define the category of modules (also called algebras) over it by considering the objects $x \in \mathcal{C}$ with an action map $\rho \colon Mx\rightarrow x$ satisfying the same relations as for the algebraic counterpart.
In commutative algebra for a fixed ring $R$ we can define left $R$-modules and right $R$-modules. Moreover, we can define the opposite ring $R^*$ with multiplication $a\ast b=ba$, it follows immediately that $(R^*)^*\cong R$ and left $R^*$-modules coincide with right $R$-modules (and vice versa right $R^*$-modules are left $R$-modules).
We can put together two compatible actions of $R^*$ and $R$ and obtain a $R$-bimodule. That is, we require an abelian group $N$ to have a left and a right $R$-action such that $a(nb)=(an)b$ for all $a,b \in R$ and $x \in N$. This is equivalent to $N$ having the structure of left $R^e$-module where $R^e=R\otimes_{\mathbb{Z}} R^*$.
My question is the following: can we transpose the notions of left/right modules and bimodules also for a categorical monad? From what I illustrated I would think that we need only to define for a generic monad $M$ its opposite $M^*$ in sensible way, but I have never seen such construction anywhere.