For a monoid with zero $M$, how many additive operations on $M$ can there be making $M$ a ring? Let $(M,\times)$ be a monoid with zero. Let $\Sigma(M,\times)$ be the set of binary operations $+$ on $M$ such that $(M,+,\times)$ is a ring. Let $\sim$ be an equivalence relation on $\Sigma(M,\times)$ defined by $$+_1\sim+_2\iff(M,+_1,\times)\cong(M,+_2,\times).$$
Let's denote the quotient set $\Sigma(M,\times)/\sim$ by $\Sigma'(M,\times)$ and consider the number $$\mathrm{add}(M,\times)=|\Sigma'(M,\times)|.$$
I have several questions about the behavior of this function, none of which I know how to approach.
$(1)$ For an integer $n\geq 0$, is there always a monoid $(M,\times)$ such that $n=\mathrm{add}(M,\times)?$
$(2)$ The same question with the requirement that $M$ be finite.
$(3)$ The two previous questions are equivalent if $\mathrm{add}(M,\times)\geq\aleph_0$ for $M$ infinite. Is it true? It is false by Todd Trimble's answer. $(\mathbb Z,+)$ with a zero element adjoined is another example. 
$(4)$ Is there an upper bound to the values of $\mathrm{add}(M,\times)$ over all finite monoids with $0?$ What about all monoids with zero?
 A: As for question 3, if the question is whether it's true that there are always countably many nonisomorphic additive group structures that extend a structure of infinite monoid with zero to a ring structure, the answer is no. In fact, there need not be any such extensions. 
For example, consider a monoid given by a meet-semilattice with a top and bottom element; the multiplication is defined to be the meet. Every element is idempotent and so any ring extension would have to be a Boolean ring. There is in fact at most one way that a poset can be a Boolean algebra (so the number of ring extensions is either zero or one), but most posets are not Boolean algebras. For example, for $n \geq 2$ the poset of linear subspaces of $\mathbb{R}^n$ cannot be a Boolean algebra, because the distributive law fails. 
A: I think the answer to (2) is yes (which also answers (1) and (4), of course).
First, note that $R=\mathbb{Z}/4\mathbb{Z}$ and $S=\mathbb{F}_2[x]/(x^2)$ have isomorphic multiplicative monoids (via the map $0\mapsto 0$, $1\mapsto 1$, $2\mapsto x$, $3\mapsto 1+x$).
Call this monoid $M$.
Then the direct product $M^n$ of $n$ copies of $M$ is the multiplicative monoid of
$T_k=R^k\times S^{n-k}$ for $k=0,\dots,n$, giving $n+1$ rings that are pairwise non-isomorphic, since they have different additive groups.
I claim that these are the only rings with $M^n$ as multiplicative monoid. Suppose $T$ is another. Then it follows just from the multiplicative structure that $T$ is commutative and has $n$ primitive idempotents $e_1,\dots,e_n$ such that, as a ring, $T\cong Te_1\times\dots\times Te_n$, where each $Te_i$ is a ring whose multiplicative monoid is $M$.
But it's quite easy to see that $R$ and $S$ are the only rings with multiplicative monoid $M$: the subring generated by $1$ can only be $\mathbb{Z}/4\mathbb{Z}$ or $\mathbb{Z}/2\mathbb{Z}$, so either the ring is isomorphic to $R$ or is a two-dimensional $\mathbb{F}_2$-algebra generated by an element $x$ with square zero, in which case it must be isomorphic to $S$. 
