$R$ is a ring. $\forall 0\ne a\in R \exists ! b\in R; aba=a$. Can I conclude that $R$ is a ring with identity? How? $R$ is a ring, containing more than one element, such that 
 $\forall 0\ne a\in R \exists ! b\in R; aba=a$.
Here's what I did in the very first place:
$(aba)a^{-1}=aa^-1 \text{ and } a^{-1}(aba)=a^{-1}a\Longrightarrow \left\{ \begin{array}{rl} ab=1 & \forall0\ne a\in R \exists !b\in R \\ ba=1 & \forall0\ne a\in R \exists !b\in R \end{array} \right.$
Therefore, since $ab\in R\Longrightarrow 1\in R$. So $R$ is a ring with identity.

But then I realized, how one can say $aa^{-1}=1$ when one doesn't know whether $R$ is a ring with identity.
 A: Try to first show $R$ has no zero divisors (see below), and having that, it follows that $bab=b$.
With this in hand, let $c\in R$. Then $aba=a$ implies $caba=ca$, so cancelling the nonzero $a$, $cab=c$. 
On the other hand, $bab=b$ implies $bc=babc$, so cancelling the nonzero $b$ yields $c=abc=cab$. That is, $c=(ab)c=c(ab)$ for any $c\in R$. So $ab$ satisfies the properties of $1$, so $ab=1$. 

To see there are no zero divisors, assume $a\neq 0$. Let $b$ be the unique $b$ such that $aba=a$. Suppose $c$ is such that either $ca=0$ or $ac=0$. Then 
$a(b+c)a=aba+aca=a$ so by uniqueness, $b=b+c$, or $c=0$.
A: $a = aba = abababa = ... = (ab)^n a \quad \forall n$
Therefore $a=(ab)a$ and $a=a(bab)a$.
But $\exists ! b$. So $b=bab$. So if $a\mapsto b$ by that rule, $b\mapsto a$.
Now $ab$=$abab$=$(ab)^n \forall n$
So $\forall c \in R, \quad ab c=(ab)^2c=(ab)^n c$
If $c \neq (ab)c$, then since $(ab)\neq 0$ 
$(ab)c\neq (ab)^2 c$. But this is false, so $c=(ab)c$ for all $c$, therefore $ab=1$
A: aba = a
aba - a = a - a
aba - a = 0
EDIT:
aba - aba = 0
baba - baba = 0
baba - ba = 0
a(bab - b) = 0
b = bab
Now since b = bab let c be any element in R
since bc = babc we see that c = (ab)c
we also have ca = caba
thus c = c(ab)
Therefore ab must be a an identity in R
