For all $a, b \in R $ with $a \neq 0$, the equation $ax = b$ has a solution in$R$ Suppose $R $  is a non zero ring.

How can we prove the following conditions are equivalent? 

In Proof 2 to 3, how can we prove  from $ a R = R $  that the ring has identity?


  1:  For all $a, b \in  R $ with $a \neq 0$, the equation $ax = b$ has a solution in$R$ .
  
  2: $ R^{2} \neq 0$  and $ R$ has no right ideals other than $ 0$ and $R$.
  
  3: $ R$ is a division ring.
The set of all elements $x$ such that $xR=0$ forms a right ideal of $R$. 

By hypothesis it can't be all of $R$, so it is just $\{0\}$.

In other words, $xR=R$ for nonzero $x$.

Let $x$ be a nonzero element. Two observations:


The right annihilator of $x$ is a right ideal, and it can't be $R$, so it is zero. Therefore $x$ is left cancelable. (Notice this is true for any nonzero $x$.)
$xe=x$ for some $e$, and furthermore $xee=xe=x$. Canceling $x$ we find that $e^2=e$.


Using the $e$ above, we see that $ey=e^2y$ implies $y=ey$ for any $y\in R$, so $e$ is a left identity.

Finally, there's probably a shorter trick, but this one is the first that occurred to me:
$(y-ye)^2=y^2-y^2e-yey+yeye=y^2-y^2e-y^2+y^2e=0=(y-ye)0$. If $y-ye\neq0$, we could cancel it on the left, obtaining $y-ye=0$, a contradiction. Therefore $y-ye=0$, and $e$ is a right identity too.
If $aR = R$, $a$ is not zero, then $R$ has identity:
we recall from rschwieb that left cancellation is valid. Now:

$aR = R \implies ac = a$ for some $c \in R$

$cR = R \implies cb = a$ for some $b \in R$

Prove that $a = b$ and then $c$ will be our identity:
$ab = acb = a(cb) = aa \implies b = a$,
$ac = cb = ca = a \implies c$ is the identity.