Answer
See below
Work Step by Step
Assume $A=\begin{bmatrix}
a & b \\ c & d
\end{bmatrix} \in S$
We can write set $S$ as $S=\{A \in M_2(R):a+b+c+d=0\}$
We have $a+b+c+d=0 \rightarrow a=-b-c-d$
Thus, $\begin{bmatrix}
a& b \\ c & d
\end{bmatrix}=\begin{bmatrix}
-b-c-d& b \\ c & d
\end{bmatrix}=b\begin{bmatrix}
-1 & 1 \\ 0 & 0
\end{bmatrix}+c\begin{bmatrix}
-1 & 0\\ 1 & 0
\end{bmatrix}+d\begin{bmatrix}
-1 & 0 \\ 0 & 1
\end{bmatrix}$
We can see that every matrix $A \in S$ is a scalar multiple of the matrix $\begin{bmatrix}
-1 & 1 \\ 0 & 0
\end{bmatrix},\begin{bmatrix}
-1 & 0\\ 1 & 0
\end{bmatrix},\begin{bmatrix}
-1 & 0 \\ 0 & 1
\end{bmatrix}$
Hence, the set $\{\begin{bmatrix}
-1 & 1 \\ 0 & 0
\end{bmatrix},\begin{bmatrix}
-1 & 0\\ 1 & 0
\end{bmatrix},\begin{bmatrix}
-1 & 0 \\ 0 & 1
\end{bmatrix}\}$ spans $S$
