Equality of left and right inverses in a finite domain

I'm attempting to understand the proof of the statement that every finite domain $R$ is a division ring. For a fixed nonzero element $a \in R$, the left-translation map is injective since $R$ is a domain. Since $R$ is finite this map is also surjective which implies that 1 is in its image. Therefore we have the existence of an element $b \in R$ such that $ab=1$, right inverses exist for nonzero elements. By looking at the right-translation map, we get the existence of left inverses.

Given a finite domain $R$ and $a,b,c \in R$ such that $ab=1$ and $ca=1$ how do we show that $b=c=a^{-1}$? How do we know that the nonzero elements are units?

I think the answer is as follows: The nonzero elements already form a multiplicative monoid since $R$ is a domain. But to form a group it is enough to demonstrate the existence of left (or right) inverses only by these alternate group axioms.

So maybe we just need left inverses and an identity to get a division ring.

In any ring $R$, if $a$ has both a left and a right inverse, then the left and right inverses are the same and the element is a unit.

To see this in your case, just write this: $$ b = 1\cdot b = (ca)b = c(ab) = c\cdot 1 = c.$$ Since $b=c$, then $ca = ba = ab = 1$, so $b=c=a^{-1}$, as desired.

This is the standard argument. It can be used, for example, to show that a function that has left and right inverses must be invertible (and the left and right inverses are the same).

It is also true that if you have a ring without zero divisors, which has a left identity $e\neq 0$ and every element has a left inverse, then the ring is a division ring. The argument is the one you give, essentially showing that the nonzero elements form a group under multiplication. But the fact that an element having both left and right inverses is in fact invertible is more basic than that.

What happens if you mutliply the identity $ab = 1$ on the left by $c$?