I'll write $\to$ instead of $\supset$, and $\bot$ instead of false, below.
Since Law of Excluded Middle is given, I'll argue using classical propositional logic.

Since $M\to\bigcirc M$ is already given as Axiom $\bigcirc$R, let's prove $$\bigcirc M\to M.$$
We are given
$$\neg\bigcirc\bot.\tag{*}$$
We need to show
$$\neg M\to\neg\bigcirc M,$$
in other words
$$A\to\neg\bigcirc\neg A.$$
So it suffices to show
$$A\wedge\bigcirc\neg A\to\bot$$
so by (*) it suffices to obtain
$$A\wedge\bigcirc\neg A\to \bigcirc\bot.$$
But we have by Axiom $\bigcirc R$ and Axiom $\bigcirc S$,
$$A\wedge\bigcirc\neg A\implies \bigcirc A\wedge\bigcirc\neg A\implies\bigcirc (A\wedge\neg A)\iff \bigcirc\bot$$
so we're done.

>Note that Axiom $\bigcirc M$ was not needed.