I will assume your regions are closed and have differentiable boundary, otherwsie reflections are not defined for certain directions.

## Question 1: a no-go result.

The answer to Question 1 is no: the reflection property you are interested in is symmetric in $A$ and $B$.

For every ray $v$ originating at $A$, write $f_A(v)$ for the (unique, by convexity) point where $v$ hits the boundary of the convex region. Because of convexity, the map $f$ is a bijection between the unit circle and the boundary of the region. If $v$ is now a ray originating at $B$, similarly define $f_B(v)$ to be the unique point of the boundary hit by a ray from $B$ in direction $v$.
If $x$ is a point on the boundary, write $n_x$ for the normal to the boundary at that point.

By hypothesis, every ray through $A$ reflects through $B$ after hitting the boundary once. This is the same as saying that for every $x$ on the boundary the directions $f_A^{-1}(x)$ and $f_B^{-1}(x)$ form the same angle to $n_x$. The statement is symmetric in $A$ and $B$, so your condition holds for all reflections from $A$ to $B$ if and only if it holds for all reflections from $B$ to $A$.

## Question 2: a no-go result.

The answer to Question 2 is also no: at some point the two reflections must collapse to one.

Indeed, consider any convex region and let $f_A(v)$, $f_B(v)$ and $n_x$ as before. Because the boundary is closed and differentiable and the region is convex, there must exist an $x$ on the boundary such that $n_x = -f_A^{-1}(x)$, i.e. a ray from $A$ towards $x$ reflects back through $A$. Either $B$ is on the way of the reflection, in which case we're done, or the ray must pass through $A$, hit the boundary again at another point $y$ and from there reflect through $B$. But then the ray from $A$ in direction $f_A^{-1}(y)$ reflects at $y$ and then goes through $B$ after a single reflection.

## Question 2: relaxing assumptions.

The answer to Question 2 can become yes if some assumptions are relaxed. In the example(s) below, we must allow a finite number of directions in which the second focus is hit only after a single reflection (because of the the no-go result above).

**Cofocal parabolas**

For this first example we have to allow the points A and B to coincide (must as the circle is a degenerate case of an ellipse where the foci coincide). Then the answer is yes: joining two parabolic arcs with the same focus and axis of symmetry (but opposite vertices wrt the focus) always yields a convex shape with the desired property. Indeed, every ray from the focus (point A) will reflect exactly twice before reaching the focus again (point B), with the exception of two rays parallel to the axis of symmetry of the parabolas. Note that the two rays orthogonal to the axis of symmetry of the parabolas hit the boundary in a non-differentiable point, but the reflection is nevertheless well-defined by continuation.

**Half-ellipse and two half-circles**

For this second example we have to allow non-convex shapes and we have to relax the requirement of two reflections: let's allow the ray from A to travel through B after the first reflection, as long as it always goes through B after the second reflection.

Consider two distinct points A and B in the plane and let $a > 0$ be their distance. Without loss of generality, let the points be at $(\pm a/2, 0)$ in the Cartesian plane. Draw the positive y-coordinate half of an ellipse with foci A and B and having semimajor axis $a$ (so that the semiminor axis $b$ satisfies $\frac{a}{2} = \sqrt{a^2-b^2}$, i.e. $b = \frac{\sqrt{3}a}{2}$):
$$
y = \frac{\sqrt{3}}{2}\sqrt{a^2-x^2}
$$
Draw the negative y-coordinate half of a circle around $A$ with diameter $a$:
$$
y = -\sqrt{\frac{a^2}{4}-\left(x+\frac{a}{2}\right)^2}
$$
Draw the negative y-coordinate half of a circle around $B$ with diameter $a$:
$$
y = -\sqrt{\frac{a^2}{4}-\left(x-\frac{a}{2}\right)^2}
$$

A ray from A with positive y direction will first reflect against the ellipse boundary, then reflect against the circle boundary below B and finally go through B (but it will have passed through B on its way from the ellipse boundary to the circle boundary).
A ray from A with negative y direction will first reflect against the circle boundary below A, then reflect against the ellipse boundary and finally go through B.

## Bonus: additional "reflections" of a physical nature.

The answer to Question 1 is related to a property of classical geometric optics sometimes known as reciprocity: the path described by light is independent of the direction in which light travels through the path. Specifically, if a ray starting at $A$ in direction $v$ follows a path which ends at $B$ in direction $w$, then light starting at $B$ in direction $-w$ follows the same path (travelling in the opposite direction) and arrives at $A$ in direction $-v$.

It may interest you to know that this property *fails* in curved spacetimes: a ray starting at $A$ in direction $v$ may follow a path which ends at $B$ in direction $w$, but light starting at $B$ in direction $-w$ might not even reach $A$ at all. As a pictoresque (and extreme) example of this, consider light falling radially into a Schwarzschild black hole: rays starting a point $A$ outside directed radially towards a point $B$ inside will reach $B$, but rays starting from $B$ inside directed radially towards the point $A$ outside will never reach $A$.

guessis that a candidate C needs to be smooth for it to have a chance of achieving the above properties. $\endgroup$