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.