Burnside convolution Let $G$ be a group. Say that an orbit is a nonempty transitive $G$-set. Let $\Xi$ be a set of finite orbits such that each finite orbit is isomorphic to exactly one element of $\Xi$.
If $X,Y,Z\in\Xi$, then $X\times Y$ is a disjoint union of orbits, and the multiplicity of $Z$ in $X\times Y$ is a well-defined nonnegative integer $b(X,Y,Z)$. 
Say that $G$ satisfies Condition (C) if for all $Z\in\Xi$ there are only finitely many pairs $(X,Y)\in\Xi^2$ such that $b(X,Y,Z)\ne0$. 

Question 1. Do all groups satisfy Condition (C)?

The motivation for introducing Condition (C) is that, if it holds, then we can define a convolution on the group $A:=\mathbb Z^\Xi$ of all maps from $\Xi$ to $\mathbb Z$ by 
$$
(f*g)(Z):=\sum_{X,Y\in\Xi}f(X)\ g(Y)\ b(X,Y,Z),
$$ 
and it is easy to see that $(A,+,*)$ is a commutative ring with one, which coincides with the Burnside ring of $G$ if $G$ is finite. 
(We can of course take only the finitely supported fonctions in $\mathbb Z^\Xi$, but this was the subject of this question.
If we start with a monoid $M$ instead of a group $G$, we can generalize the above lines by replacing the notion of orbit by that of nonempty $M$-set which is not a disjoint union of nonempty sub-$M$-sets, and we can ask 

Question 2. Do all monoids satisfy Condition (C)?

The monoids given as examples in this question satisfy Condition (C), but I don't even know if the additive monoid $\mathbb N$ does, so let me ask formally 

Question 3. Does $\mathbb N$ satisfy Condition (C)?

Edit. In view of this answer
it seems appropriate to ask a fourth question.
Say that an $M$-set is indecomposable if it is neither empty nor a disjoint union of nonempty sub-$M$-sets. 
Say also that $M$ satisfies Condition (D) if for all $X,Y,Z$ such that 
$\bullet\ X$ and $Y$ are two finite indecomposable $M$-sets, 
$\bullet\ Z$ is a maximal indecomposable sub-$M$-set of the product $X\times Y$, 
the map $Z\to X$ induced by the projection is surjective.
Clearly groups satisfy Condition (D). 
Note that (D) implies (C), because, up to isomorphism, there are only finitely many quotients of a given finite $M$-set.

Question 4. Do all monoids satisfy Condition (D)?

 A: Let me restate your question for groups. Transitive $G$-sets modulo isomorphisms are in bijective correspondence with the set of subgroups modulo conjugation, and finite ones correspond to finite index subgroups. For subgroups $A,B,C$ of $G$, the condition $b(G/A,G/B,G/C)\neq 0$ means that there exist conjugates $A',B'$ of $A,B$ such that $C=A'\cap B'$. 
Since every finite index subgroup is contained in only finitely many subgroups, this leaves finitely many possibilities for $A',B'$, and hence, up to conjugation, finitely many possibilities for $A,B$.
So the answer (to question 1) is yes.
A: Edit 2. The answer all questions is Yes: see Edit 1 below. End of Edit 2.

The answer to Question 3 is Yes.
Say that an $M$-set is indecomposable if it is neither empty nor a disjoint union of nonempty sub-$M$-sets. 
Say also that $M$ satisfies Condition (D) if for all $X,Y,Z$ such that 
$\bullet\ X$ and $Y$ are two finite indecomposable $M$-sets, 
$\bullet\ Z$ is a maximal indecomposable sub-$M$-set of the product $X\times Y$, 
the map $Z\to X$ induced by the projection is surjective.
Clearly groups satisfy Condition (D). 
Note that (D) implies (C), because, up to isomorphism, there are only finitely many quotients of a given finite $M$-set.
So it suffices to show that $\mathbb N$ satisfies (D).
For ease of notation replace the additive monoid $\mathbb N$ with the multiplicative monoid $M$ freely generated by the element $a$. 
Say that a point of an $M$-set is periodic if it is a fixed point of $a^n$ for some $n\ge1$. 
The proof of the following facts is left to the reader:
(a) If $y$ is a periodic point of an $M$-set $X$ and $n$ is a nonnegative integer, then $y=a^nx$ for some $x\in X$.
(b) If $x$ is a point of a finite $M$-set, then $a^nx$ is periodic for $n$ large enough.
(c) If $x$ and $y$ are two points of a finite indecomposable $M$-set, and if $y$ is periodic, then there is an $n\in\mathbb N$ such that $a^nx=y$.
Let us prove that $M$ (or equivalently  $\mathbb N$) satisfies (D).
Let $(x_0,y_0)$ be in $Z$ and let $x$ be in $X$. It is enough to show that there is a $y\in Y$ such that $(x,y)\in Z$. By (b) there is an $i\in\mathbb N$ such that $(x_1,y_1):=a^i(x_0,y_0)$ is a periodic point in $Z$. There is, by (c), a $j\in\mathbb N$ such that $a^jx=x_1$, and, by (a), a $y\in Y$ such that $a^jy=y_1$. This implies $a^j(x,y)=(x_1,y_1)\in Z$, and thus $(x,y)\in Z$, as desired.

Edit 1. Let us prove that the answer to Question 4 is Yes. This will imply that the answer all questions is Yes.
The argument is similar to the one used above to answer Question 3, but I prefer to make this edit self-contained. 
Let us fix an element $a$ of $M$. Say that a point of an $M$-set is periodic if it is a fixed point of $a^n$ for some $n\ge1$. 
The proof of the following facts is left to the reader:
(a) If $y$ is a periodic point of an $M$-set $X$ and $n$ is a nonnegative integer, then $y=a^nx$ for some $x\in X$.
(b) If $x$ is a point of a finite $M$-set, then $a^nx$ is periodic for $n$ large enough.
In the setting of Question 4, let $p:X\times Y\to X$ be the projection, and assume by contradiction that $p(Z)$ is a proper subset of $X$. Then there is a tuple $(a,x_1,x_2,y_2)$ with 
$$
a\in M;\ x_1,x_2\in X;\ x_1\notin p(Z);\ ax_1=x_2;\ y_2\in Y;\ (x_2,y_2)\in Z.
$$ 
It suffices to show $x_1\in p(Z)$. By (b) we can pick an $n\in\mathbb N$ such that $a^n(x_2,y_2)\in Z$ is periodic. Set 
$$
x_3:=a^nx_2=a^{n+1}x_1,\ y_3:=a^ny_2.
$$ 
By (a) there is a $y_1\in Y$ such that $a^{n+1}y_1=y_3$, and we get 
$$
a^{n+1}(x_1,y_1)=(x_3,y_3)\in Z,\ 
$$ 
which implies $(x_1,y_1)\in Z$ and thus $x_1\in p(Z)$, contradiction.
