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 $$\neg\bigcirc\bot\tag{*}$$ $$\bigcirc\bigcirc M\to\bigcirc M\tag{$\dagger$}$$ $$\bigcirc M\wedge\bigcirc N\to\bigcirc (M\wedge N)$$ 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 $$A\wedge\bigcirc\neg A\implies \bigcirc A\wedge\bigcirc\neg A\implies\bigcirc (A\wedge\neg A)\iff \bigcirc(\bot)$$ so we're done. I guess I didn't use ($\dagger$).