The answers to the question Can every true theorem that has a proof be proven by contradiction? show that if a theorem can be proven directly, it can be proven by contradiction.
The main arguments presented there are of the form:
Let P be a proof of the theorem. Assume the theorem is false. However we can exhibit P, which contradicts the assumption
However this argument assumes that we can rely on the validity of the direct proof in our proof by contradiction.
My question is then, if you have a theorem that can be proven directly, is it possible to prove the theorem by contradiction, but without using the fact that there exists a direct proof in your proof by contradiction?
Not necessarily -- but for even less interesting reasons than the argument you don't like!
The following is fact:
The proof is very simple -- since the claim to be proved is one of the axioms of PA, simply stating it with "(axiom)" next to it constitutes a proof.
On the other hand, the other axioms of PA are not sufficient to prove the claim -- to see this, observe that we get a model of those other axioms by taking the universe to be $\mathbb N$, $S$ to be the successor function, $+$ to be addition, and $\cdot$ to be the constant function that always returns $0$. This model does not satisfy the claim we want to prove, so since first-order logic is sound, the claim does not have a proof.
In other words: Every proof of $\forall x\forall y( x\cdot Sy = x\cdot y + x )$ in PA must at some point prove $\forall x\forall y( x\cdot Sy = x\cdot y + x )$ directly -- because that is the only way the proof can depend on that axiom, as it must.