The answer is no in general (but it's not difficult to check that the answer is yes for injective operators).
Counterexample. Let $H$ be a Hilbert space and $T_0: H \to H$ a bounded linear operator with non-closed range. Consider the operator $T: H \times H \to H \times H$ given by $T(x,y) = (T_0 x, 0)$ for all $x,y \in H$, where $H \times H$ is endowed with, say, the usual norm induced by $H$ that turns $H \times H$ into a Hilbert space. The range of $T$ is $(T_0 H) \times \{0\}$, so it is not closed.
However, let us now show that $T(S_{H \times H}) = T_0(B_H) \times \{0\}$, where $B_H$ denotes the closed unit ball in $H$. The inclusion "$\subseteq$" is clear. For the converse inclusion "$\supseteq$", take $x \in B_H$. Then one can choose $y \in B_H$ such that $(x,y) \in S_{H \times H}$ and hence $(T_0 x, 0) = T(x,y) \in T(S_{H \times H})$.
Since $H$ is reflexive, the set $B_X$ is weakly compact. As $T_0$ is continuous with respect to the weak topology, it follows that $T_0(B_X)$ is also weakly compact, hence weakly closed, and thus closed. Therefore, $T(S_{H \times H}) = T_0(B_H) \times \{0\}$ is closed in $H \times H$.
Remark. Of course, a similar construction can also be done with other reflexive Banach spaces.