We could also play this game with groups. One starts with a group $G$. A move consists in replacing $G$ by $G/\langle\langle a \rangle\rangle$, i.e. we mod out the smallest normal subgroup containing $a \neq 1$. The ending condition holds iff the ascending chain condition with respect to normal subgroups holds (do these groups have a name?). When $G$ is abelian, this means that $G$ is finitely generated.

Actually we can play this game for *every algebraic structure*: Given a variety in the sense of universal algebra, start with with an algebra $A$. A move consists in replacing $A$ by $A/a \sim b$, where $a,b \in A$ with $a \neq b$. The game proposed by Will Sawin is the game on rings, including the zero ring, under the *misère play rule*, i.e. the last one moving loses (see Tom Goodwillie's comment).

I have tried to analyze this game for abelian groups, non-abelian groups, and rings in the article Algebraic games. There are lots of scattered examples, but for abelian groups the structure theorem makes it possible to give a general answer which ones are $\mathcal{P}$ (i.e. are losing positions):

Let $A$ be a finitely generated abelian group.

Under the normal play rule, $A$ is a $\mathcal{P}$-position if and only if $A$ is a square, i.e. $A \cong B^2$ for some finitely generated abelian group $B$.

Under the misère play rule, $A$ is a $\mathcal{P}$-position if and only if $A$ is either a square, but not elementary abelian of even dimension, or elementary abelian of odd dimension.

Now for something on-topic, some results about the game on rings:

If $R$ is $\mathcal{P}$, then $\mathrm{Spec}(R)$ is connected (Lemma 6.2). Let $R$ be a Dedekind domain. If $R$ has some principal maximal ideal, then $R$ is $\mathcal{N}$. Otherwise, $R$ is $\mathcal{P}$ (Prop. 6.3). It follows, for example, that $k[x,y]/(y^2-x^3+x-1)$ is $\mathcal{P}$. Hence, $k[x,y]$ is $\mathcal{N}$. Section 6.2 is devoted to zero-dimensional rings (whose complexity was already mentioned by Tom Goodwillie), finally showing that the cusp $k[x,y]/(y^2-x^3)$ is $\mathcal{P}$.

Many problems remain open, for example if $k[x_1,\dotsc,x_n]$ is $\mathcal{N}$ for all $n \geq 1$.