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.