My **main point** is that I think a sufficient condition is to have an axis of symmetry. 

Before I get to that, a few other remarks.

1) On the left below is a variation of the right triangle construction showing that a convex (or simply not too badly reentrant) polygon with a right angle has a simple periodic billiard path (spbp) of the form $abcdcba.$ There are arbitrary further sides not shown. 

One might require that a spbp touch all the sides in order to avoid arbitrary modifications which don't affect the path, but for some examples below it is convenient not to require this.

2) The top middle diagram is an isosceles triangle with spbp $bacab$ and another $defgfed.$

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 I'll now mainly restrict to sppbp with the second (or first) p standing for *polygonal*.

3) A comment gives a construction for an sppbp in an isosceles triangle $PQP'$. I've illustrated it (as I understand it) for the case that the triangle is acute, I'm not sure about the obtuse case.  From $P'$ draw the perpendicular to side $PQ$ hitting that side at $b$ The path starts at the center $a$ of the base and goes to $b.$ The claim (which I do not dispute) is that it continues parallel to the base and creates a, sppbp $abb'a.$

**main point** 

Perhaps continuity arguments are enough to show that a convex polygon with a line of symmetry has a sppbp (maybe several). 

3) (again) We certainly have the degenerate path $aQa.$ For construction 3 can't we consider a path starting from the center $a$ of the base and hitting $PQ$ at a variable point $b?$ There is a choice of $b$ , as in 2. , starting a path $abaca.$ If we move $b$ slightly closer to $Q$ then the path would return to the base a little to the right od $a.$ If we move it close to $Q$ then the next part of the path hits side $P'Q$ even closer to $Q.$ someplace in between is a choice which continues parallel to the basegiving $abb'a.$


4) The final picture is a convex hexagon $PQRR'Q'P'$ with line of symmetry $ad$ (so $a$ and $d$ are midpoints of their sides) but no other special structure.

The degenerate path $ada$ is only so interesting. However a simple billiard path starting at $a$ and getting to $d$ without crossing the lie $ad$ will continue on to an sppbp.  I've put one in (rather freehand ) which starts $abcd.$ Can we argue by continuity (moving $b$ along edge $PQ$) that there must be such?

If we move $b$ closer to $Q$ then there should be a place where one get a pentagonal sppbp

Remove sides $QR$ and $Q'R'$ the extend the other sides to create a trapezoid. That should allow a quadrilateral sppbp which would also be valid for the hexagon. 

Using every other side and extending gives an isosceles triangle. This allows at least some of the paths from 2. and 3.

This reasoning is a little unsubstantiated, but seems reasonable for any convex $2k$-gon with a line of symmetry. 

The case of an odd number of sides should not be much different. 


![enter image description here][1]


  [1]: https://i.sstatic.net/fZl5Y.png