Is there an easy characterization of all sets $M \subseteq \mathbb{R}^2$ with the following property?
A point $(x_M,y_M)$ (which may depend on $M$) exists such that each straight line through $(x_M,y_M)$ cuts $M$ into two halves of equal Lebesgue measure.
Let's use a polar coordinate system with $M$ as the origin. Then the property can be stated as: the area of $M$ within the angular range $\alpha\le \theta\le \alpha+\pi$ is independent of $\alpha$. This area is $$\frac12 \int_\alpha^{\alpha+\pi} \int_0^\infty \chi_M(r,\theta) r\,dr\,d\theta$$ where $\chi_M$ is the characteristic function of $M$. Differentiation with respect to $\alpha$ yields $$I(\alpha) =I(\alpha+\pi) \quad \text{for a.e. } \alpha, \text{ where }\ I(\alpha) = \int_0^\infty \chi_M(r,\alpha) r\,dr$$ This is a necessary and sufficient condition. It doesn't look particularly geometric, but that's what it is.
In the special case when the region is star-shaped about $M$, i.e., is described by $r\le f(\theta)$, the condition becomes $f(\theta)=f(\theta+\pi)$ a.e., which is central symmetry (up to a set of zero measure).
A non-centrally symmetric example: annulus-type domain $$ \sqrt{1+\cos\theta}\le r\le \sqrt{2+\cos\theta} $$ pictured below:
Of course, $\sqrt{f(\theta)}\le r\le \sqrt{A+f(\theta)}$ works as well, for any function $f$ and any constant $A$.