Prove that S is a subring of $M_2(\mathbb{R})$

Say I have a ring $M_2(\mathbb{R})$, which is the ring of $2x2$ matrices with real entries.

Say I also know that $$S = \left\{\begin{pmatrix} a & b\\ c & d \end{pmatrix} \mid a,b,c,d \in \mathbb{R}, a + b = c+ d\right\}.$$

I'm trying to prove that $S$ is a subring of $M_2(\mathbb{R}).$

I assume that this is a matter of testing for $\begin{pmatrix} 0 & 0\\ 0 & 0 \end{pmatrix}$ and $\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}$ in $S$ and then looking at addition and multiplication of matrices.

Am I on the right track? Is there a different way I should be looking at the proof?

If $I$ is the identity, $U=\begin{pmatrix}1&0\\1&0\end{pmatrix}$ and $V=\begin{pmatrix}0&1\\0&1\end{pmatrix},$ then if $a+b=c+d,$ $$\begin{pmatrix}a&b\\c&d\end{pmatrix}=(a+b-c)I+bV+cU$$

Show $S$ is exactly all $\alpha I+\beta U+\gamma V,$ with $\alpha,\beta,\gamma\in \mathbb R.$

Then use $UV=V^2=V, VU=U^2=U$ to prove $S$ is closed under multiplication.

Also show $I,\mathbf 0\in S.$

Alternatively, let $v_0=\begin{pmatrix}1\\1\end{pmatrix}.$ Then $A\in S$ if an only if, for some $\lambda\in \mathbb R,$ $Av_0=\lambda v_0.$

Then if $A_1, A_2$ have eigenvector $v_0$ with eigenvalues $\lambda_1,\lambda_2$ then show $$(A_1+A_2)v_0=(\lambda_1+\lambda_2)v_0\\ (A_1A_2)v_0=(\lambda_1\lambda_2)v_0.$$

Also, show $I$ and $\mathbf 0$ are in $S.$

This latter approach works in general. If $v_0\in\mathbb R^n$ is a fixed vector, define:

$$S_{v_0}=\left\{A\in M_n(\mathbb R)\mid \exists\lambda\in\mathbb R(Av_0=\lambda v_0)\right\}$$

$S_{v_0}$ is always a subring.

If $v_0\neq 0,$ the $\lambda$ associated for any $A\in S_{v_0}$ is unique, and the map $A\to\lambda_A$ is a homomorphism $S_{v_0}\to \mathbb R.$

The first technique is harder to generalize.