Solution:

Since $A E$ bisects $\angle R A M$, we have $R E = E M$, and $E, A$ lie on different sides of $R M$. Since $A M$ is a diameter, $\angle A R M = 90^{\circ}$. If the midpoint of $R M$ is $N$, then from $[R A M] = [R E M]$ and $\angle A R M = 90^{\circ}$, we find $A R = N E$. Note that $O$, the center of $\omega$, $N$, and $E$ are collinear, and by similarity of triangles $N O M$ and $R A M$, $O N = \frac{1}{2} A R = \frac{1}{2} N E$. Therefore, $O N = \frac{1}{3}$ and $N E = \frac{2}{3}$. By the Pythagorean theorem on triangle $R A M$, $R M = \frac{4 \sqrt{2}}{3}$. Therefore, the area of $M A R E$ is $2 \cdot \frac{1}{2} \cdot \frac{4 \sqrt{2}}{3} \cdot \frac{2}{3} = \frac{8 \sqrt{2}}{9}$.