Answer
See below
Work Step by Step
Assume $A=\begin{bmatrix}
a & b & c \\ d & e &f \\g & h & i
\end{bmatrix} \in M_3(R):a+b+c=d+e+f=g+h+i=a+d+g=b+e+h=c+f+i=0$
First, we have the system:
$a+b+c=0\\d+e+f=0\\g+h+i=0 \\
\rightarrow a=-b-c\\
d=-e-f\\g=-h-i$
Substitute: $\begin{bmatrix}
a & b & c \\ d & e &f \\g & h & i
\end{bmatrix}=\begin{bmatrix}
-b-c & b & c \\ -e-f & e &f \\ -h-i & h & i
\end{bmatrix}$
Then $b+e+h=0\\c+f+i=0 \\ (-b-c)+(-f-e)+(-h-i)=0\\
\rightarrow b=-e-h\\
c=-f-i\\-b-c=e+f+h+i$
Thus $\begin{bmatrix}
e+f+h+i & -e-h & -f-i \\ -e-f & e &f \\ -h-i & h & i
\end{bmatrix}$
We can see that every matrix $A \in S$ is a scalar multiple of the matrix $\begin{bmatrix}
e+f+h+i & -e-h & -f-i \\ -e-f & e &f \\ -h-i & h & i
\end{bmatrix}=e\begin{bmatrix}
1 & -1 & 0\\ -1 & 1 & 0 \\ 0 & 0 & 0
\end{bmatrix}+f\begin{bmatrix}
1 & 0 & -1\\ -1 & 0 & 1 \\ 0 & 0 & 0
\end{bmatrix}+h\begin{bmatrix}
1 & -1 & 0\\ 0 & 0 & 0 \\ -1 & 1 & 0
\end{bmatrix}+i\begin{bmatrix}
1 & 0 & -1\\ 0 & 0 & 0 \\ -1 & 0 & 1
\end{bmatrix}$
Hence, the set $\{\begin{bmatrix}
1 & -1 & 0\\ -1 & 1 & 0 \\ 0 & 0 & 0
\end{bmatrix},\begin{bmatrix}
1 & 0 & -1\\ -1 & 0 & 1 \\ 0 & 0 & 0
\end{bmatrix},\begin{bmatrix}
1 & -1 & 0\\ 0 & 0 & 0 \\ -1 & 1 & 0
\end{bmatrix},\begin{bmatrix}
1 & 0 & -1\\ 0 & 0 & 0 \\ -1 & 0 & 1
\end{bmatrix}\}$ spans $S$
