Yes, this is indeed a subspace.
To see that it is a subspace we need to check that it is closed under scalar multiplication and addition.
Choose any scalar $z\in \mathbb{C}$ then $$ z \left( \begin{array}{ccc}
a & a & a \\
0 & 0 & a \\
a & a & a \end{array} \right) = \left( \begin{array}{ccc}
za & za & za \\
0 & 0 & za \\
za & za & za \end{array} \right) \in S$$
and for any $a, b\in \mathbb{C}$ we have, $$ \left( \begin{array}{ccc}
a & a & a \\
0 & 0 & a \\
a & a & a \end{array}\right) + \left( \begin{array}{ccc}
b & b & b \\
0 & 0 & b \\
b & b & b \end{array} \right) = \left( \begin{array}{ccc}
a+b & a+b & a+b \\
0 & 0 & a+b \\
a+b & a+b & a+b \end{array} \right) \in S$$
Thus we see that $S$ is closed under scalar multiplication and addition of vectors, so it is a subspace.