If, in a unit-regular ring, the right annihilator of $a$ equals the right annihilator of $b$, then $aR = bR$?

Question

Recall that a (unital) ring $R$ is von Neumann regular (VNR) if, for each $x \in R$, there exists $y \in R$ such that $x = xyx$; and unit-regular if such an element $y$ can be taken to be a unit.

Question. Assume $R$ is a VNR (resp., unit-regular) ring, and let $a, b \in R$ be such that the right annihilator $r(a)$ of $a$ equals to the right annihilator $r(b)$ of $b$. Does it follow that $aR = bR$?

It is relatively simple to see that, if $R$ is a VNR ring and $r(a) = r(b)$ for some $a, b \in R$, then $Ra = Rb$: This is true, more generally, for a right principally injective ring (see, e.g., Lemma 5.1 in Nicholson and Yousif's book on quasi-Frobenius rings); and it is a matter of fact that VNR rings are (left and) right principally injective.

Answer 1

This seems false if $r(a)=\{x\in R\mid ax=0\}$. The ring $M_n(F)$ of $n\times n$ matrices over a field $F$ is a counterexample for $n\geq 2$. Two matrices have the same right annihilator iff they have the same null space and $aM_n(F)=bM_n(F)$ if and only if they have the same image or column space. It is easy to find matrices with the same null space but different column space like take any non-zero vector $v$ and make all rows $v$ versus making the first row $v$ and all other rows zero. It is well known all semisimple rings are unit regular.