Showing a ring where $ax = b$ has a solution for all non-zero $a, b$ is a division ring 
Let $R$ be a ring. Suppose that for all $a, b \in R$ with $a, b$ not equal to $0$, the equation $ax = b$ has a solution. Show that $R$ is a division ring.

I know that I need to show that $R$ has a multiplicative identity and inverse; however, I really have no idea of how to start. 
 A: The statement is false for the $1$-element ring. It satisfies the
condition, but it is not a division ring acccording to the
standard definition. But the statement is true for rings
of more than one element.
As a hint on this problem,
I will explain why $R$ must have an identity element.
Assume that $|R|>1$.
For $a\in R$ let $\lambda_a\colon R\to R\colon x\mapsto ax$
denote the function that is left multiplication by $a$.
It is an endomorphism of the underlying additive group of $R$.
The condition that 
for all $a, b\in R-\{0\}$ the equation $ax=b$ has a solution
expresses exactly that $\lambda_a$ is surjective when $a\neq 0$.
Claim. $\lambda_a$ is injective when $a\neq 0$.
Assume otherwise that $a\neq 0$,
$c\in R-\{0\}$ and $\lambda_a(c)=ac = 0$.
Since $\lambda_a$ is surjective, for any $b\in R-\{0\}$ there exists
$d\in R-\{0\}$ such that $ad=b$. Since $\lambda_c$ is surjective,
there exists $e\in R-\{0\}$ such that $ce=d$. Now we have
$0 = 0e = (ac)e = a(ce) = ad = b \neq 0$, a contradiction. \\\
Thus $\lambda_a$ is a permutation of $R$ for every $a\neq 0$.
In particular, there is a unique element $u_a\in R$ such that
$\lambda_a(u_a) = a$; i.e., $au_a = a$. For any $r\in R$ we have
$u_ar - r\in\ker(\lambda_a) = 0$, so $u_ar = r$ holds for any $r\in R$
and any $a\in R-\{0\}$. This says that $u_a$ is a left unit of $R$
when $a\neq 0$. Moreover, $u_a$ acts like a right unit
on the singleton set $\{a\}$, i.e., $au_a=a$.
Claim. If $a, b\in R-\{0\}$, then $u_a=u_b$.
If not, then $c:=u_a-u_b$ is not zero, 
hence $\lambda_c$ is a permutation. Now for $d\neq 0$
we have $0\neq \lambda_c(d) = (u_a-u_b)d = u_ad-u_bd = d-d = 0$,
since the $u$-elements are left units. Contradiction. \\\
So far we have that each element of the form $u_a$ acts
as a left identity on $R$ and a right identity on $\{a\}$,
But they are all the same $u$,
which is therefore a 2-sided identity on $R$.
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