On page 264 , 2nd edition. Theorem 5.1 It says Let M be a free $R$-module with "a basis" $\{e_1,\dots,e_n\}$ Then $M$ is $R$-isomorphic to $R^n$. Above he is defining the standard basis as the $n$-tuples $\{e_1,\dots,e_n\}$ witch is only one abelian group.
Does he really mean that this works only if M has this particular basis? I think this should be valid for any basis of any free module. There are a lot more sets that can be made into abelian groups then the $n$-tuples.
The example above states that any infinite cyclic group as a $\mathbb Z$-module has a basis, which contradicts his notation below according to me.
I don't think he means the "standard basis" in the hypothesis of the proof even though he has written it.
Having a basis $B$ (or in other words being a free module) means that the module behaves exactly like $R^{|B|}$, up to isomorphism. So you can do all of your thinking while having $R^{|B|}$ with the standard basis in mind, and then use the assumed isomorphism to argue that all your conclusions are valid for arbitrary free modules too.
For example, this means that any infinite cyclic group is $\mathbb Z^1$ (that is, just $\mathbb Z$ itself) up to isomorphism.