Does there exist a division ring without unity? In abstract algebra I have only ever seen division introduced via multiplicative inverses, namely starting from a ring with unity $R$ and then adding the condition that each element $x$ has an inverse element $x^{-1}$ such that $xx^{-1}=x^{-1}x=1$. But I can also imagine a concept of division without having a unit element, defined as follows:
Let $R$ be a ring with the property that for each ordered pair $(a,b)\in R$ with $b\neq 0$, there exists a unique $c\in R$ such that $a=bc=cb$. Therefore it makes sense to define $a/b:=c$, where $c$ is the unique element corresponding to $(a,b)$ as specified above. Is it possible for such a structure to exist on a ring without unity?

Let $R$ be a ring without unit.  Suppose

for each ordered pair $(a,b)\in R$ with $b\neq 0$, there exists a unique $c\in R$ such that $a=bc=cb$.    

We claim that $R$ is a ring with unit.
Let $b \in R$, $b \ne 0$.  Then there is a unique $e_b$ such that $b = be_b = e_bb$.  
We must show that $e_a = e_b$ for all nonzero $a,b$.  Then this will be the unit in $R$.  
Let $a,b \in R$, both nonzero.  There is $c$ so that $a = cb = bc$.  So
$$
e_b a = e_b b c = b c = a,\qquad
ae_b= c b e_b = c b = a.
$$
Thus, $e_b$ satisfies the defining property of $e_a$.  By the uniqueness, $e_a = e_b$.
