Let $R$ be a ring without unity. Let $x \in R$ be such that there exists a $y \in R$ with
$$-x-y+xy=0=-x-y+yx \hspace{0.1cm}.$$
I have to prove that $x$ in quasi-regular, i.e. there exists a $z \in R$ such that
$$x+z+xz=0=z+x+zx \hspace{0.1cm}.$$
Any ideas?
This isn't even true for rings with identity:
For example, in $\mathbb R$, with $x=-1$ you have $y=1/2$ satisfying $x+y=xy=yx$.
But you cannot find $z$ such that $-1+z=z$.
Now if you still want to shoehorn this into a ring with no identity, then you just take $S=(\mathbb R,+,\circ)$ where $\circ$ is zero for every product, and form the product ring $\mathbb R\times S$, and use $x'=(-1,0)$, $y'=(1/2,0)$ and show no $z'=(r_1,r_2)$ works.