<s>[INST] Why do we assume $a+b=b+a$ in a Ring with 1. Also, is it true with Rings without 1? I was wondering why do some people use redundant axioms in definitions?
If you just expand $(a+1)(b+1)=(a+1)b+a+1=ab+b+a+1$ $(a+1)(b+1)=a(b+1)+b+1=ab+a+b+1$. Hence, $ab+a+b+1=ab+b+a+1$, then cancel ab and 1. Then, you get it's commutative for free. 
Why do we then assume it? 
Also, is it even stronger. If we drop the condition that 1 is in R, then can we still deduce a+b=b+a? As clearly the argument just done falls apart if you haven't got 1 in R. 
 [/INST] Without a $1$, you cannot deduce the condition. Take $(R,\cdot,\times)$ where $(R,\cdot)$ is any nonabelian group with identity element $e$, and let $a\times b = e$ for all $a$ and $b$ (the "zero multiplication ring"). This satisfies all the axioms of a ring without $1$, except for commutativity of the first operation. So $a+b=b+a$ cannot be deduced from the rest of the axioms of a ring without a $1$.
Another method to deduce commutativity when there is a $1$ is to consider $(a+b)(1+1)$ and distribute both ways: 
$$\begin{align*}
(a+b)(1+1) &= (a+b)+(a+b) = a+b+a+b,\\
(a+b)(1+1) &= a(1+1) + b(1+1) = a+a+b+b.
\end{align*}$$
From $a+b+a+b = a+a+b+b$ we deduce $b+a=a+b$.
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