Looking for a reference of quasi inverse in rings This question is related to exercise 1.51 from Rotman's "Introduction to the Theory of Groups". 
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My question is related to exercise 1.51 from Rotman's "Introduction to the Theory of Groups".
An element a
in a ring $R$ (with unit element $1$) has a left quasi-inverse if there exists an element $b\in R$ such that $a+b−ba=0$. I want to show that if every element in $R$ has a left quasi-inverse except $1$, then $R\setminus{1}$ is a group under the operation $a∗b=a+b−ba$.
If you take a ring and change its multiplication by the new, this new structure has anything interesting?
Is some reference papers or books about this operation?
 A: First I'll answer your question about the exercise (showing that $(R\setminus \{1\},*)$ is a group).
Let's start by checking that $*$ is associative on all of $R$: 
\begin{align*}
(a * b)*c &= (a+b-ba)*c\\
&= (a+b-ba)+c-c(a+b-ba)\\
&= a+b+c-ba-ca-cb+cba\\
&= a+(b+c-cb)-(b+c-cb)a\\
&= a*(b+c-cb)\\
&= a*(b*c).
\end{align*}
Now observe that for any $a\in R$:
\begin{align*}
 1*a &= 1+a-a1 = 1\\
 a*0 &= a+0-0a = a\\
 0*a &= 0+a-a0 = a
\end{align*}
And if $b$ is a left quasi-inverse of $a$, then $a*b = 0$.
Now we'd like to show that $R\setminus \{1\}$ is closed under $*$. Let $a,b\in R\setminus \{1\}$, and suppose for contradiction that $a*b = 1$. Since $b\neq 1$, $b$ has a left quasi-inverse $c$, so 
\begin{align*}
(a*b)*c &= 1*c\\
a*(b*c) &= 1\\
a*0 &= 1\\
a &= 1
\end{align*}
which is a contradiction.
So now we have an associate operation $*$ on $R\setminus \{1\}$ which has an identity element $0$, and every element has a right-inverse (for all $a$ there exists $b$ such that $a*b = 0$). So you can apply the fact that a monoid with right-inverses is a group. (Proof: Let $b$ be the right-inverse of $a$, and let $c$ be the right-inverse of $b$. Then $(a*b)*c = 0*c$, so $a*(b*c) = c$, so $a = c$, so $b*a = 0$, and $b$ is the inverse of $a$). 

Ok, having solved the exercise, some comments:
An element with a left quasi-inverse is called left quasiregular. For a reference, you could start with Wikipedia and follow the references there. 
An element $x$ is left quasiregular if and only if $(1-x)$ has a left inverse. Indeed, if $y(1-x) = 1$, then the left quasi-inverse of $x$ is $z = -xy$. We have $$x+z-zx = x-xy+xyx = x(1-y+yx) = x(1-y(1-x)) = x(1-1) = 0.$$
Conversely, if $x+z-zx = 0$, then $y=(1-z)$ is the left inverse of $(1-x)$. We have $$y(1-x) = (1-z)(1-x) = 1-x-z+zx = 1-(x+z-zx) = 1.$$
Suppose $R$ has the property that every element of $R\setminus \{1\}$ is left quasiregular, as in the exercise. Then for all $x\neq 1$, $(1-x)$ has a left inverse. Thus every nonzero element of $R$ has a left inverse. This means that $R\setminus \{0\}$ is a monoid with left-inverses, and hence a group (by the dual of the fact noted above). So every element of $R\setminus \{1\}$ is left quasiregular if and only if $R$ is a division ring.
