Suppose $|S| \ge 2$. For any $p \in S$, let $R(p)$ denote the stable rectangle with upper-right corner $p$. We say such $p$ is pivotal if $p + (1, 1) \notin S$ and $|R(p)|$ is even.

Claim — If $|S| \ge 2$, then a pivotal $p$ always exists.
Proof. Consider the top row of $S$.
* If it has length at least 2, one of the two rightmost points in it is pivotal.
* Otherwise, the top row has length 1. Now either the top point or the point below it (which exists as $|S| \ge 2$) is pivotal. $\square$
We describe how to complete the induction, given some pivotal $p \in S$. There is a partition
$$S = R(p) \sqcup S_1 \sqcup S_2$$
where $S_1$ and $S_2$ are the sets of points in $S$ above and to the right of $p$ (possibly empty).
Claim — The desired inequality holds for stable subsets containing $p$.
Proof. Let $E_1$ denote the number of even stable subsets of $S_1$; denote $E_2$, $O_1$, $O_2$ analogously. The stable subsets containing $p$ are exactly $R(p) \sqcup T_1 \sqcup T_2$, where $T_1 \subseteq S_1$ and $T_2 \subseteq S_2$ are stable.
Since $|R(p)|$ is even, exactly $E_1E_2 + O_1O_2$ stable subsets containing $p$ are even, and exactly $E_1O_2 + E_2O_1$ are odd. As $E_1 \ge O_1$ and $E_2 \ge O_2$ by inductive hypothesis, we obtain $E_1E_2 + O_1O_2 \ge E_1O_2 + E_2O_1$ as desired. $\square$
By the inductive hypothesis, the desired inequality also holds for stable subsets not containing $p$, so we are done.