They are the same statement, but about different numbers: one is about n', the other one is about n, which corresponds to S n' in the goal. Induction is precisely about showing that something is valid for S n' just by knowing that it holds for n'. The fact that they are different numbers is what makes this non-trivial. For example, it is not possible to show that S n' is odd assuming that n' is odd, even though both statements are "the same".
You asked about the mechanics of rewriting. After simplification, your goal looks like this:
n' : nat
IHn' : n' = n' + O
---------------------------------------
S n' = S (n' + O)
When you call rewrite, you are telling Coq to replace n' + O by n' in the goal. This leads to the statement S n' = S n', which is a simple equality that can be discharged by the axiom of reflexivity.
Edit
Here is another way to think about induction. Let's forget about the induction tactic for a second. It is possible to prove that 0 = 0 + 0:
Lemma plus_0_0 : 0 = 0 + 0.
Proof. reflexivity. Qed.
We can prove similar statements about any concrete natural number:
Lemma plus_1_0 : 1 = 1 + 0.
Proof. reflexivity. Qed.
Lemma plus_2_0 : 2 = 2 + 0.
Proof. reflexivity. Qed.
However, we can obtain these proofs differently, making fewer assumptions about the shape of the number we are talking about. First, we prove the following conditional statement.
Lemma plus_Sn_0 : forall n', n' = n' + 0 -> S n' = S n' + 0.
Proof. intros n' Hn'. simpl. rewrite <- Hn'. reflexivity. Qed.
This lemma states the inductive step of a proof by induction. Note that it does not say that n = n + 0 holds for all natural numbers n, but that it holds only for numbers of the form n = S n' provided that n' = n' + 0. We know that this works for n' = 0, because we proved it above. Combined with plus_Sn_0, we have another proof of 1 = 1 + 0.
Lemma plus_1_0' : 1 = 1 + 0.
Proof. apply plus_Sn_0. apply plus_0_0. Qed.
Now, we know that the statement holds for n' = 1, we can play the same trick for n = 2:
Lemma plus_2_0' : 2 = 2 + 0.
Proof. apply plus_Sn_0. apply plus_1_0'. Qed.
This proof does not use directly the fact that 0 = 0 + 0. This is why the fact that we proved the base case is irrelevant for proving the inductive step: all we need to know about is the predecessor of the number we are interested in -- in this case, this predecessor being 1.
Naturally, we could have expanded out the proof of plus_1_0' inside the proof of plus_2_0':
Lemma plus_2_0'' : 2 = 2 + 0.
Proof. apply plus_Sn_0. apply plus_Sn_0. apply plus_0_0. Qed.
More generally, given any natural number constant n, we can prove that n = n + 0 without induction by applying plus_Sn_0 n times followed by plus_0_0. The principle of mathematical induction extrapolates this intuitive observation for any expression n denoting a natural number, and not just constants.
