First note that your proposed $\sigma$-ring is not a $\sigma$-ring: $\{ 0 \} \in \mathcal{R}$, but $\varnothing = \{ 0 \} \setminus \{ 0 \} \notin \mathcal{R}$.
Note that since $\bigcap_{n=1} E_n = E_1 \setminus \bigcup_{n=2} ( E_1 \setminus E_n )$ it follows that $\mathcal{R}$ is closed under intersections of nonempty countable subfamilies.
We will also make use of the following set identity: $$F \cup E = ( E^\text{c} \setminus F )^\text{c}.$$
Note, first, that $\mathcal{M}$ is clearly closed under complements, so it suffices to show that it is closed under countable unions. If $\mathcal{A}$ is a countable subfamily of $\mathcal{M}$, let $\mathcal{A}^+ := \mathcal{A} \cap \mathcal{R}$, and $\mathcal{A}^- := \mathcal{A} \setminus \mathcal{R}$. As $\mathcal{A}^+$ is a countable subfamily of $\mathcal{R}$, then $A := \bigcup \mathcal{A}^+ \in \mathcal{R}$. If $\mathcal{A}^- = \varnothing$, there is nothing else to do, so we assume that $\mathcal{A}^- \neq \varnothing$.
Using the above set identity (and de Morgan's laws) it follows that
$$\begin{multline}
{\textstyle \bigcup} \mathcal{A}
= {\textstyle \bigcup} \mathcal{A}^+ \cup {\textstyle \bigcup} \mathcal{A}^-
= A \cup {\textstyle \bigcup_{E \in \mathcal{A}^-}} E =
%\\ =
{\textstyle \bigcup_{E \in \mathcal{A}^-}} ( A \cup E )
% = {\textstyle \bigcup_{E \in \mathcal{A}^-}} ( E^\text{c} \setminus A )^\text{c}
= ( {\textstyle \bigcap_{E \in \mathcal{A}^-}} ( E^\text{c} \setminus A ) )^\text{c}.
\end{multline}$$
Since $E^\text{c} \in \mathcal{R}$ for each $E \in \mathcal{A}^-$, the closure properties of $\mathcal{R}$ imply that $\bigcap_{E \in \mathcal{A}^-} ( E^\text{c} \setminus A ) \in \mathcal{R}$, and so $\bigcup \mathcal{A} \in \mathcal{M}$.
