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.

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 $B_H$ is weakly compact. As $T_0$ is continuous with respect to the weak topology, it follows that $T_0(B_H)$ 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.