If I am understanding the question correctly, you want x^2=0 in this ring. If that is the case, a typical element of the ring is a+bx, and we want to show that there exist c+dx so that
a+bx = (a+bx)(c+dx)(a+bx)
It is then trivial to check that (c+dx) = (a-bx)/a^2 works, so every element with a!=0 can be "inverted". However, it is clear that since x^2=0, if a=0 then the element can't be "inverted", and hence the ring is not von Neumann regular.