Yes: the condition is equivalent to : any straight line through $p$ intersects $S$ in a segment whose midpoint is $p$, that is the convex is center-symmetric. $$*$$
Details: Assume $p=0\in\mathbb{C}$ and let $H _ +$ and $H _ -$ be resp. the upper and lower half-planes. Consider the difference of the areas of the two halves of $S$ cut by a rotating line: $$\delta(\alpha):=\operatorname {Area}(S\cap e^{i\alpha}H _ +) - \operatorname {Area}(S\cap e^{i\alpha}H _ -) $$ The derivative of this quantity wrto $\alpha$ is $$\delta'(\alpha)=\frac{1}{2}\operatorname {Length}(S\cap e^{i\alpha}\mathbb{R} _ -) - \frac{1}{2}\operatorname {Length}(S\cap e^{i\alpha}\mathbb{R} _ +), $$ as it immediately follows writing the areas as integrals in polar coordinates. In particular, $\delta(\alpha)$ is constant iff $p$ is the midpoint of $S\cap e^{i\alpha}\mathbb{R}$.
Remark: the same argument holds of course if we only assume $S$ to be convex with respect to $p$ ("star-shaped"). If we also drop the condition of star-shapeness, it is easy to make non-symmetric counterexamples: e.g. the set of all $z$ with $|z|\le 4$ in the upper half-plane and all $z$ with $3\le |z|\le 5$ in the lower half-plane.