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 $$\neg\bigcirc\bot.\tag{*}$$ First, by Axiom $\bigcirc R$, $$\neg M\wedge\bigcirc M\to\bigcirc \neg M\wedge\bigcirc M.$$ Therefore by $\bigcirc S$, $$\neg M\wedge\bigcirc M\to\bigcirc (\neg M\wedge M)$$ Therefore by definition of $\bot$, $$\neg M\wedge\bigcirc M\to\bigcirc\bot$$ By (*), $$\neg(\neg M\wedge\bigcirc M)$$ By de Morgan and law of excluded middle, $$M\vee \neg\bigcirc M$$ So, $$\bigcirc M\to M$$
Note that Axiom $\bigcirc M$ was not needed.