Let $R$ be a ring with every element but $1$ having a left quasi-inverse. Then $R-\{1\}$ is a group under $a*b=a+b-ba$. This 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 $R - \{1\}$ is a group under the operation $a*b=a+b-ba$.
What I have an issue with is showing closure, i.e. that $a+b-ba=1$ iff $a=1$ or $b=1$. I noticed that $x+y-yx=1$ is equivalent to $(1-y)(x-1)=0$, but this is as far as I can get, since we don't know that the ring has no zero-divisors. 
 A: The $*$ operation is associative and has $0$ as neutral element (direct verification). Moreover, the left quasi-inverse of $a$ ($a\ne1$) cannot be $1$, because $a+1-1a=1\ne0$ (the $0\ne1$ assumption must be made, of course, or $R\setminus\{1\}$ would be empty and so not a group).
If $c$ and $d$ are left-quasi inverses of $a$ and $b$ respectively, you have
$$
(d*c)*(a*b)=(d*(c*a))*b=(d*0)*b=d*b=0.
$$
Therefore $a*b$ has a left quasi-inverse and the operation is well-defined on $R\setminus\{1\}$. Actually this shows that $a*b$ belongs to the set of elements having a left quasi-inverse, without assuming that all elements (except $1$ which can't) have a left quasi-inverse.
Now just show that a left quasi-inverse of a left quasi-inverse is …
A: Under the bijection $\phi: R\setminus\{1\}\to R\setminus\{0\}$, $x\mapsto 1-x$, we note that $*$ becomes $\cdot$, i.e. $a*b=c\iff (1-a)(1-b)=1-c$. Therefore, the existence of a quasi-inverse for $a\in R\setminus\{1\}$ is equivalent to the existence of a multiplicative inverse for $1-a$. Thus if all elements of $R\setminus\{1\}$ have a quasi-inverse, we see that every element of $R\setminus\{0\}$ is a unit and $R\setminus\{0\}$ is just the group $R^\times$ of units under multiplication. Consequently, $\phi$ is a group isomorphism between $(R\setminus\{1\},*)$ and $(R^\times,\cdot)$.
