Probably these results will serve to resolve your question
- In an arbitrary ring $R$, if $rad(R)\neq R$ then $rad(R)=\{r\in R| \forall x,y\in R: \;xry\mbox{ is right quasi-regular}\}$
proof: Remember that $Rad(M)=\bigcap\{M|M\mbox{ maximal modular right ideal of }R\}$ and let $H=\{r\in R| \forall x,y\in R: \;xry\mbox{ is right quasi-regular}\}$. Note that if an element $a\in R$ is not a right quasi-regular then there is a modular maximal ideal $M$ such that $a\notin M$. in this way, if $a$ is not right quasi-regular, the right ideal $S=\{at-t| t\in R\}$ does not contain $a$.
In particular, applying the Zorn Lemma you can see that there exist a right ideal $M$ maximal in the set of all the right ideals which contains $S$ but not the element $a$. moreover, it is easy to verify that this ideal is a right maximal in $R$. On the other hand $M$ is modular since $ar-r\in M$ for each $r\in R$, so if $b\notin H$ there are elements $x$, $y$ in $R$ such that $xby$ is not right quasi-regular. then there is a maximal modular right ideal which does not contain $xby$ and hence, does not contain $b$, and $b\notin Rad(R)$.
Conversely, one has to prove first that if $M$ is a maximal modular right ideal in $R$ then $H\subseteq (R:M)=\{s\in R| \forall m\in M, sm\in R\}$ (the proof of this propiety you can found in T.Y.Lam, a first course of noncommutative ring or every book in ring theory). Since $rad(R)\neq R$ then there exists a maximal modular right ideal $M$ in $R$ and $H\subseteq \bigcap\{(R:M)| M\mbox{maximal modular right ideals in }R\}$. Hence $RH\subseteq M$ for each $M$ and $M$ being modular, we also have $H\subseteq M$. Finally, $H\subseteq rad(R)$.
- In a ring with identity $R$, $rad(R)=\{r\in R|\forall s\in R\mbox{ the element }1-rs \mbox{ is right invertible}\}$.
proof: It's time to use your observation, you see that in a ring with identity $R$ an element $r\in R$ is right quasi-regular if and only if is right invertible ($r+s-rs=0\Leftrightarrow (1-r)(1-s)=1$). Now, we define
$M=\{r\in R|xry \mbox{ is right quasi-regular in }R,\forall x\in R\}$ and $N=\{r\in R|xr \mbox{ is right quasi-regular in }R,\forall x\in R\}$, because the ring having identity, then $M \subseteq N$. On the other hand if $xr+t-xrt=0$ then $xry+ty-xrty=(xr+t-xrt)y=0$ and so $N\subset M$, and you can use the previous result, so the identity is done.