The answer you're not hoping to hear probably is: for any ring with 1, the (right) simple modules are characterized as quotients R/M where M is a maximal right ideal. So, if you can determine all the maximal right ideals, you have all the simple right modules, and similarly for the left side. I'm sure you can immediately find several, but I'll have to apologize I can't remember if there's a slick way to point out all the simple modules at once.
It may also be helpful to know that the radical is the set of all strictly upper triangular matrices (since all the maximal left/right ideals will have to contain these). I think selecting any diagonal entry and looking at the subset of matrices zero on that diagonal entry is a maximal ideal. I thought this might be a complete set, but then I read that not all simple modules embed into an upper triangular matrix ring, so there must be a few more hiding in there.
I know that's pretty lackluster, so I'll tell you everything else I know about this ring. It's an Artinian serial ring (this means it's a direct sum of right ideals, each of whom has a finite linearly ordered set of submodules). It's also hereditary, and is a ring of finite representation type.
I look forward to seeing a more confident and complete solution for this problem :)