$\mathbb N$ will denote the set $\{0,1,2,\ldots\}.$ The semigroup $(\mathbb N,+)$ doesn't have a zero element. $\mathbb N^0$ will denote the semigroup $\mathbb N$ with zero adjoined, that is the set $\mathbb N\cup\{\star\},\;\star\not\in\mathbb N,$ with the operation $+_0$ defined by $$a+_0b=\begin{cases}a+b & \text{ for } \{a,b\}\subset\mathbb N\\\star &\text{ for } \{a,b\}\not\subset\mathbb N\end{cases}$$
Is there a ring whose multiplicative structure is isomorphic to $\mathbb N^0?$ I believe there isn't, but I don't see a proof.
I've proved that the analogously defined semigroup $\mathbb Z^0$ isn't isomorphic to the multiplicative structure of any ring, but I could do that because such a ring would have to be a field, which made the task easier. Such a field would have to be of characteristic $2$ because the multiplicative group of any other field contains $\{-1,1\}$ as a finite subgroup isomorphic to $\mathbb Z/2\mathbb Z,$ and $(\mathbb Z,+)$ doesn't have non-trivial finite subgroups. Therefore, the field would have to be an extension of $\mathbb F_2.$ If it were an algebraic extension, its multiplicative group would either be finite or contain a non-trivial finite subgroup, which is again impossible. However, a transcendental extension would have to have a copy of $\mathbb F_2(x)$ in it, whose multiplicative group is not finitely generated. Every subgroup of $\mathbb Z$ is finitely generated.
The characteristic restriction carries over to the question I'm asking. If $1\neq -1$ in a ring, then its group of units has $\mathbb Z/2\mathbb Z$ as a subgroup, which $\mathbb N^0$ doesn't. So any ring whose multiplicative structure is isomorphic to $\mathbb N^0$ must be of characteristic $2.$ Even more, it would have to have a trivial group of units, as there are no invertible elements in $\mathbb N^0$ other than $0.$ It must obviously be a countable commutative ring with unity. There are rings satisfying these conditions: $\mathbb F_2[x]$ does. However, the multiplicative structure of $\mathbb F_2[x]$ isn't isomorphic to $\mathbb N^0.$ (The proof of this I've found seems slightly too complicated to me, so if you have a very simple one, I'd be very glad to see it. Mine involves counting "prime" elements in the semigroups, defined like for rings.) If I could reduce this question to $\mathbb F_2[x],$ similarly to the case of $\mathbb Z$ and $\mathbb F_2(x),$ it would be great but I don't see how to do it.