Multisets are containers, also called bags. A multiset is a set that can have repeats:

$$\{ a, a, a, c, b, c \}$$

Usually when researchers talk about multisets, they use this kind of presentation:

$$\psi = 3| a \rangle + 2| c \rangle + | b \rangle$$

In fact, $$\psi$$ is an element of an $$\mathbb{N}$$-module. I had thought that $$\mathbb{N}$$-modules are the Eilenberg-Moore category of the multiset monad. I now think that's not the case. There is no notion of addition of the scalar multiples in the Eilenberg-Moore category of the multiset monad. Instead, that category is just commutative monoids. This means that there is an intermediary monad, $$\mathcal{MMeas}$$, which is a measures monad where you can add the multiplicities. In that monad, the functor takes a set to multiplicities over set elements like this:

$$F: \{ a, b, c \} \rightarrow \{ \{ 3|a \rangle, 1|b \rangle , 2 |c \rangle \}, \ldots\}$$

$$U: Multiset \rightarrow \mathcal{MMeas}$$

is an equivalence and that is why multisets are always written like this:

$$\psi = 3| a \rangle + 2| c \rangle + | b \rangle$$

It is this monad whose Eilenberg-Moore category is $$\mathbb{N}$$-modules.

Is this true or am I overthinking this?

I had thought that $$\newcommand{\N}{\mathbf{N}}\newcommand{\Z}{\mathbf{Z}}\N$$-modules are the Eilenberg-Moore category of the multiset monad. I now think that's not the case. […] Instead, that category is just commutative monoids.”
But $$\N$$-modules are precisely commutative monoids. This just like the better-known fact that $$\Z$$-modules are precisely Abelian groups: every commutative monoid carries a unique $$\N$$-module structure (this is easy check directly, or it can be abstractly seen by viewing an $$\N$$-module structure on $$A$$ as a semiring map $$\N \to \mathrm{End}(A)$$ and remembering that $$\N$$ is the initial semiring), so the forgetful functor from $$\N$$-modules to comm monoids is an isomorphism of categories.
So if you’re happy that the E-M category of the multiset monad is commutative monoids, then that shows that it’s $$\N$$-modules. Concretely, the addition comes from the multiplication of the multiset monad: writing $$\newcommand{\M}{\mathcal{M}}$$ for the monad, we have $$\M1 \cong \N$$, and under this isomorphism, the monad multiplication $$\mu : \M^2 1 \to \M 1$$ corresponds to addition on $$\N$$, seen as a map $$\M \N \to \N$$.