Is the multiplicative identity unique in a unit ring? In my coursebooks and on various websites online (wiki, proofwiki, etc.), among the first theorems which follow the definition of a ring are the uniqueness of the additive identity and the additive inverse. But I haven't found an answer to the following:

Question: If $R$ is a ring with multiplicative identity, is the multiplicative identity unique?

I suspect it is the case, since by definition, $1\cdot r=r\cdot 1=r$ for all $r\in R$; so if we assume that $1$ and $1'$ are two distinct multiplicative identities in $R$, we must have $1=1\cdot 1'=1'$, i.e. $1=1'$ — a contradiction. Is my line of reasoning correct? If so, why is this not included with all the simple theorems? 
 A: You are correct, and $1 = 1 \cdot 1' = 1' \Rightarrow 1 = 1'$ is a great proof by contradiction.
As for "not included with all the simple theorems", it has probably already been presented in your book for far "simpler" groups and been treated as already-known facts by this point.
A: 
Is my line of reasoning correct? 

Yes, it appears many places on this website, too. Actually you don't even need to frame it as a contradiction (the advice is usually to use direct proofs, where possible.) You simply say, "suppose $1$ and $1'$ are identities. Then $1=11'=1'$. Thus there is only a single identity. QED.

If so, why is this not included with all the simple theorems?

It typically is, or is included as an easy exercise.
If you think about it, a good general version of the theorem is just "the identity element of a monoid is unique."  You only need the one operation, and the definition of an identity element. (It doesn't have anything to do with inverses. You could even relax the operation to be nonassociative.)
