Sure, we can define such things. Let's work in the Morita 2-category $\text{Mor}(k)$ over a commutative ring $k$, which has
- objects $k$-algebras $A$,
- morphisms $k$-bimodules (with composition given by tensor product), and
- 2-morphisms bimodule homomorphisms.
Equivalently, applying the forgetful functor $\text{Hom}(k, -)$, we can think of the Morita 2-category as having
- objects the cocomplete $k$-linear categories $\text{Mod}(A)$ of right modules over $k$-algebras,
- morphisms given by tensor product with a bimodule (equivalently, cocontinuous $k$-linear functors, by the Eilenberg-Watts theorem)
- 2-morphisms natural transformations.
The Morita 2-category is a categorified version of modules; specifically it can be thought of as a 2-category of module categories over $\text{Mod}(k)$, which itself can be thought of as a categorified commutative ring. It has a tensor-hom adjunction
$$\text{Hom}(A \otimes B, C) \cong \text{Hom}(A, [B, C])$$
where $A \otimes B$ is the ordinary tensor product over $k$ and $[B, C] = B^{op} \otimes C$ is the internal hom. This adjunction says that we can identify $(A \otimes B, C)$-bimodules naturally with $(A, B^{op} \otimes C)$-bimodules. It is moreover the case that $\otimes$ really deserves to be called the tensor product in this setting, in that $\text{Mod}(A \otimes B)$ is the universal recipient of a "bilinear" (cocontinuous and $k$-linear in each variable) functor out of $\text{Mod}(A) \times \text{Mod}(B)$.
The unit of the tensor product is $\text{Mod}(k)$, so we can define a "comonoidal object" in this setting to be equipped with a comultiplication $\text{Mod}(A) \to \text{Mod}(A \otimes A)$ (that is, an $(A, A \otimes A)$-bimodule) and a counit $\text{Mod}(A) \to \text{Mod}(k)$ (that is, a left $A$-module), plus an associator and stuff like that, satisfying some axioms that look like the axioms of a monoidal category. This is the dual of a sesquialgebra structure, in the very nice sense that a structure of this sort on $A$ is exactly a sesquialgebra structure on $A^{op}$.