If you have a finite ring $R$ that is von Neumann regular, then $[0]=N(R)$ or more generally if you have a von Neumann regular ring $R$ such that each element has a power that belongs to a subsemigroup which is a group. This follows from a result on semigroups in http://arxiv.org/abs/math/0605698, see also http://arxiv.org/pdf/0709.4341.pdf. Maybe you can use some of the ideas with the ring property to get further.

If you have a finite ring $R$ that is von Neumann regular, then $[0]=N(R)$ or more generally if you have a von Neumann regular ring $R$ such that each element has a power that belongs to a subsemigroup which is a group. This follows from a result on semigroups in http://arxiv.org/abs/math/0605698, see also http://arxiv.org/pdf/0709.4341.pdf. Maybe you can use some of the ideas with the ring property to get further.