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.