I vaguely remember seeing this theorem somewhere but am having trouble finding a proof. It is difficult to search for a proof because it is difficult in general for symbolic identities and terms like "inverse", "multiplication", "ring", and "distributive" are quite general in ring theory.
I have gotten this far:
$1=\left(ab\right)^{-1}\left(ab\right)\quad1=bb^{-1}a^{-1}a$
$\left(ab\right)^{-1}\left(ab\right)=bb^{-1}a^{-1}a$
$\left(ab\right)^{-1}\left(ab\right)a^{-1}=bb^{-1}a^{-1}aa^{-1}$
$\left(ab\right)^{-1}\left(ab\right)a^{-1}=bb^{-1}a^{-1}$
$b^{-1}\left(ab\right)^{-1}\left(ab\right)a^{-1}=b^{-1}bb^{-1}a^{-1}$
$b^{-1}\left(ab\right)^{-1}\left(ab\right)a^{-1}=b^{-1}a^{-1}$
Without commutativity, I cannot simply rearrange the $\mathrm{LHS}$ to an expression where I can eliminate $(ab)$.
I may be taking the wrong approach, so any help (including starting over from the beginning) is welcome.
Let $a,b \in R$ be units with $a',b'$ being their inverses respectively. Let $1$ be the unity of the ring.
We now show that $b'a'$ is the inverse of $ab$, i.e. $(ab)(b'a') = 1 = (b'a')(ab)$.
$$\begin{array}{rcll} (ab)(b'a') &=& a(b(b'a')) & \text{associativity} \\ &=& a((bb')a') & \text{associativity} \\ &=& a((1)a') & \text{definition of }b' \\ &=& aa' & \text{definition of unity} \\ &=& 1 & \text{definition of }a' \\ \end{array}$$
$$\begin{array}{rcll} (b'a')(ab) &=& b'(a'(ab)) & \text{associativity} \\ &=& b'((a'a)b) & \text{associativity} \\ &=& b'((1)b) & \text{definition of }a' \\ &=& b'b & \text{definition of unity} \\ &=& 1 & \text{definition of }b' \\ \end{array}$$ $\blacksquare$