Diagonal matrices are an abelian group under addition, and with multiplication they become a commutative ring $(\mathcal{D},+, *)$.
More generally, the set of $n \times n$ matrices in $M_n(\mathbb{R})=\mathbb{R}^{n \times n}$ that are simultaneously diagonalized by a given eigenbasis (see Prove that simultaneously diagonalizable matrices commute) will also yield a commutative ring. I believe any such set will be isomorphic to the diagonals, since all elements are of the form $SDS^{-1}$ for fixed $S$ and any diagonal $D$.
My hypothesis: if $\mathcal{R} \subseteq M_n(\mathbb{R})$ forms a commutative ring $(\mathcal{R},+,*)$, then $\mathcal{R} \cong \mathcal{D}$. True or false?
EDIT: False, as scalar matrices $Z(M_n(\mathbb{R}))=kI_n \not \cong \mathcal{D}$. So the hypothesis should be $\mathcal{R} \cong \mathcal{D}$ OR some subring of $\mathcal{D}$. I am considering "rings" to be unital, although rng counterexamples are still interesting.
User JCAA provided an excellent counterexample. For $\alpha, a_i \in \mathbb{R}$, consider upper triangular matrices of the form
$$
\begin{bmatrix}
\alpha & a_{2} & \dots & a_{n} \\
0 & \alpha & \dots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \dots & \alpha \\
\end{bmatrix}
=
\begin{bmatrix}
\alpha I_1 & A_{1 \times (n-1)} \\
0_{(n-1) \times 1} & \alpha I_{n-1} \\
\end{bmatrix}
$$
(The right side is block matrix notation.) For distinct matrices in this set $\mathcal{U}$, we have
$$
\begin{bmatrix}
\alpha I & A \\
0 & \alpha I \\
\end{bmatrix}
\begin{bmatrix}
\beta I & A \\
0 & \beta I \\
\end{bmatrix}
=
\begin{bmatrix}
\alpha \beta I & \alpha A + \beta A \\
0 & \alpha \beta I \\
\end{bmatrix}
=
\begin{bmatrix}
\alpha \beta & (\alpha + \beta) a_{2} & \dots & (\alpha + \beta) a_{n} \\
0 & \alpha \beta & \dots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \dots & \alpha \beta \\
\end{bmatrix}
$$
So multiplication is closed and commutative (and associative, distributive since "multiplication" is just composition of linear transformations); moreover, $I_n \in \mathcal{U}$ so this becomes a unital ring. Unlike $\mathcal{D} \cong \mathbb{R} \times \dots \times \mathbb{R}$, some $u \in \mathcal{U}$ satisfy $u^2 = 0$.
Best Answer
The answer is "no". Consider the ring of matrices with first row $(0, x,y,...,z)$ and all other entries 0. This is a ring with zero product, so commutative. It is not isomorphic to a subring of $D$.