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This is more an extended comment on @MaximeRamzi's answer. His comment reminded me of one of my favorite MO questions Distinguishing combinatorial maps by their linearizations and it explains the kernel of his map. (See also the paper inspired by this question which gives more details.)

Indeed, it is shown in my answer to the above-linked question using Fitting's decompositions and Brauer's lemma that two permutation matrices are conjugate as matrices iff they are conjugate as permutations, that if $f,g$ are maps on a finite set $X$, then the corresponding linear representations are equivalent iff the essential ranges of $f$ and $g$ are equal in $A(\mathbb Z)$ (i.e., have the same cardinality and cycle structure) and $|f^n(X)|=|g^n(X)|$ for all $n\geq 1$. It follows immediately form this that the kernel of the map $A(\mathbb N)\to R(\mathbb N)$ is generated by all $[f]-[g]$ where $f,g$ are maps on the sets of the same cardinality with the same cardinality and cycle structure essential ranges and the images of $f^n$ and $g^n$ have the same cardinality for all $n\geq 1$. But this is the kernel of the map at the end of @MaximeRamzi's answer.

There is a different ring you could associate to a monoid that is the $M$-set analogue of doing $G_0$ instead of $K_0$. For groups this gives the usual Burnside ring but for monoids it is different. I strongly suspect this is what is done in the paper I linked to in the comments but I don't know.

If we do this version of things for $\mathbb N$, then finite pointed $\mathbb N$-sets are partial maps on a finite set. If you do the Grothendieck ring of the these things with pointed-direct sum and pointed direct product, you will get something complicated. But if you factor by the relations $[X]=[Y]+[X/Y]$ for invariant $Y$, then you will get $A(\mathbb Z)\times \mathbb Z$ as an abelian group and the $\mathbb Z$ copy is a two-sided ideal, generated by the pointed set $A$ consisting of the base point and one other point that is mapped to the base point. The product of cycle with $n$ vertices (plus a base point thrown in) with the $A$ is the sum of $n$ copies of $A$.

From the partial mapping view point this is that a strongly connected partial function digraph is either a single point with no edge or a cycle.

This is more an extended comment on @MaximeRamzi's answer. His comment reminded me of one of my favorite MO questions Distinguishing combinatorial maps by their linearizations and it explains the kernel of his map. (See also the paper Steinberg - Linear conjugacy inspired by this question which gives more details.)

Indeed, it is shown in my answer to the above-linked question using Fitting's decompositions and Brauer's lemma that two permutation matrices are conjugate as matrices iff they are conjugate as permutations, that if $f$, $g$ are maps on a finite set $X$, then the corresponding linear representations are equivalent iff the essential ranges of $f$ and $g$ are equal in $A(\mathbb Z)$ (i.e., have the same cardinality and cycle structure) and $\lvert f^n(X)\rvert=\lvert g^n(X)\rvert$ for all $n\geq 1$. It follows immediately from this that the kernel of the map $A(\mathbb N)\to R(\mathbb N)$ is generated by all $[f]-[g]$ where $f$, $g$ are maps on the sets of the same cardinality with the same cardinality and cycle structure essential ranges and the images of $f^n$ and $g^n$ have the same cardinality for all $n\geq 1$. But this is the kernel of the map at the end of @MaximeRamzi's answer.

There is a different ring you could associate to a monoid that is the $M$-set analogue of doing $G_0$ instead of $K_0$. For groups this gives the usual Burnside ring but for monoids it is different. I strongly suspect this is what is done in the paper Erdal and Ünlü - Semigroup Actions on Sets and the Burnside Ring I linked to in the comments but I don't know.

If we do this version of things for $\mathbb N$, then finite pointed $\mathbb N$-sets are partial maps on a finite set. If you do the Grothendieck ring of the these things with pointed-direct sum and pointed direct product, you will get something complicated. But if you factor by the relations $[X]=[Y]+[X/Y]$ for invariant $Y$, then you will get $A(\mathbb Z)\times \mathbb Z$ as an abelian group and the $\mathbb Z$ copy is a two-sided ideal, generated by the pointed set $A$ consisting of the base point and one other point that is mapped to the base point. The product of a cycle with $n$ vertices (plus a base point thrown in) with $A$ is the sum of $n$ copies of $A$.

From the partial mapping view point this is the fact that a strongly connected partial function digraph is either a single point with no edge or a cycle.

The idea is that in a $G$-set $X$ for a group $G$, if $Y\subseteq X$ is $G$-invariant, then so is $X\setminus Y$ and you can write $X=Y+X\setminus Y$ in the Burnside ring. But this doesn't work for monoids. You can have noncomplemented invariant subsets. So one work around it to define quotient of $A(M)$ by the relations that if $Y$ is an $M$-invariant subset of $X$, then $[X] = [Y]+[X/Y]$. But I think this doesn't quite work well because $X/Y$ is a pointed $M$-set. So my feeling is that one should work in the category of pointed $M$-sets and that in the associated semiring, the action consisting of just a base point should be considered $0$. The product of pointed $M$-sets mod the invariant subset of points with one or more coordinate a base point. The sum identifies base points.

The idea is that in a $G$-set $X$ for a group $G$, if $Y\subseteq X$ is $G$-invariant, then so is $X\setminus Y$ and you can write $X=Y+X\setminus Y$ in the Burnside ring. But this doesn't work for monoids. You can have noncomplemented invariant subsets. So one work around it to define quotient of $A(M)$ by the relations that if $Y$ is an $M$-invariant subset of $X$, then $[X] = [Y]+[X/Y]$. But I think this doesn't quite work well because $X/Y$ is a pointed $M$-set. So my feeling is that one should work in the category of pointed $M$-sets and that in the associated semiring, the action consisting of just a base point should be considered $0$. The product of pointed $M$-sets is the usual direct product mod the invariant subset of points with one or more coordinate a base point. The sum identifies base points.

The idea is that in a $G$-set $X$ for a group $G$, if $Y\subseteq X$ is $G$-invariant, then so is $X\setminus Y$ and you can write $X=Y+X\setminus Y$ in the Burnside ring. But this doesn't work for monoids. You can have noncomplemented invariant subsets. So one work around it to define quotient of $A(M)$ by the relations that if $Y$ is an $M$-invariant subset of $X$, then $[X] = [Y]+[X/Y]$. But I think this doesn't quite work well because $X/Y$ is a pointed $M$-set. So my feeling is that one should work in the category of pointed $M$-sets and that in the associated semiring, the action consisting of just a base point should be considered $0$.

If we do this version of things for $\mathbb N$, then finite pointed $\mathbb N$-sets are partial maps on a finite set. If you do the Grothendieck ring of the these things with pointed-direct sum and pointed direct product, you will get something complicated. But if you factor by the relations $[X]=[Y]+[X/Y]$ for invariant $Y$, then you will get $A(\mathbb Z)\times \mathbb Z$ I believe where the $\mathbb Z$ copy is generated by the pointed set consisting of the base point and one other point that is mapped to the base point. So from the partial mapping view point this is that a strongly connected partial function digraph is either a single point with no edge or a cycle.

The idea is that in a $G$-set $X$ for a group $G$, if $Y\subseteq X$ is $G$-invariant, then so is $X\setminus Y$ and you can write $X=Y+X\setminus Y$ in the Burnside ring. But this doesn't work for monoids. You can have noncomplemented invariant subsets. So one work around it to define quotient of $A(M)$ by the relations that if $Y$ is an $M$-invariant subset of $X$, then $[X] = [Y]+[X/Y]$. But I think this doesn't quite work well because $X/Y$ is a pointed $M$-set. So my feeling is that one should work in the category of pointed $M$-sets and that in the associated semiring, the action consisting of just a base point should be considered $0$. The product of pointed $M$-sets mod the invariant subset of points with one or more coordinate a base point. The sum identifies base points.

If we do this version of things for $\mathbb N$, then finite pointed $\mathbb N$-sets are partial maps on a finite set. If you do the Grothendieck ring of the these things with pointed-direct sum and pointed direct product, you will get something complicated. But if you factor by the relations $[X]=[Y]+[X/Y]$ for invariant $Y$, then you will get $A(\mathbb Z)\times \mathbb Z$ as an abelian group and the $\mathbb Z$ copy is a two-sided ideal, generated by the pointed set $A$ consisting of the base point and one other point that is mapped to the base point. The product of cycle with $n$ vertices (plus a base point thrown in) with the $A$ is the sum of $n$ copies of $A$.

From the partial mapping view point this is that a strongly connected partial function digraph is either a single point with no edge or a cycle.