The answer is *no*, in general.

For a counterexample, consider the $\ell^p$-norm on $\mathbb{R}^2$ with $p=4$, and let $x = e_1 = (1,0)$.

We first note that the vectors $e_2 = (0,1)$ and $-e_2$ do not maximize the function
\begin{align*}
f: S \ni s \mapsto \|s-x\|+\|s+x\| \in [0,\infty).
\end{align*}

Indeed, we have $f(e_2) = f(-e_2) = 2\cdot\|(1,1)\| = 2\cdot 2^{1/4}$. On the other hand, consider the vector $s_0 = 2^{-1/4} \cdot (1,1) \in S$. Then
\begin{align*}
f(s_0) = \|(2^{-1/4} - 1, 2^{-1/4})\| + \|(2^{-1/4}+1, 2^{-1/4})\| > 2\cdot 2^{1/4}
\end{align*}
(where I, admittedly, used a computer to check the latter inequality). So $f$ is neither maximized at $e_2$ nor at $-e_2$.

Hence if $y$ maximizes $f$, then the first entry of $y$ is non-zero. But $y^* \in \mathbb{R}^2$ is the pointwise product of $y$ and $|y|^{p-2}$. Thus, the first entry of $y^*$ is also non-zero, which shows that $y^*(x) \not= 0$.