If so, may I wonder if your drop rule 2 and insist on having for most vector u there exist a v so that u*v = 1. As an icing let us say having a 0 vector. That could be something.
*It doesn't rely on rule 2, it does however rely on (u*v)*w = u*(v*w). Adding rule 2 excludes the quaternions.
v * w = v * (w + 0) # Vector space property that w + 0 = w
= v * w + v * 0 # Distributive property
Another vector space property guarantees that there exists -(v * w) such that -(v * w) + (v * w) = 0. Add it to both sides
-(v * w) + v * w = -(v * w) + v * w + v * 0
0 = 0 + v * 0 # And then reduce based on the definition of -(v * w)
0 = v * 0 # Vector space property that 0 + x = x
So you have your icing :)
*It doesn't rely on rule 2, it does however rely on (u*v)*w = u*(v*w). Adding rule 2 excludes the quaternions.