The claim you have asked us to prove is not true. If PA is consistent, then by the Incompleteness Theorem there are $\Pi_1$ statements that are independent of PA, such as Con(PA), which can be seen to be $\Pi_1$ when expressed in the form "no number is the code of a proof of a contradiction in PA". Thus, if PA is consistent, then Con(PA) is a statement that is independent of PA but provable in $PA_1$, so it is a counterexample to your claim.

Perhaps a more striking counterexample would be $\neg\text{Con(PA)}$, which is independent of PA, but refutable in $\text{PA}_1$. More generally, any statement having complexity $\Sigma_1$ or $\Pi_1$ that is independent of PA will be a counterexample to your claim, since such statements are settled by $\text{PA}_1$.

Perhaps the folklore result you meant to ask about is the following?

**Theorem.** If a $\Pi_1$ statement is independent of PA, then it is true.

Proof. If a $\Pi_1$ statement $\sigma$ is independent of PA, then it is true in some model $M\models PA$. The standard model $\mathbb{N}$ is an initial segment of $M$, and since the statement $\sigma$ is $\Pi_1$, it has the form $\sigma = \forall n \varphi(n)$, where $\varphi$ has only bounded quantifiers. Since $\sigma$ holds in $M$, it holds for all standard $n$ in $M$ and hence $\sigma$ is true in the standard model. In other words, it is true. QED

Note that the proof that $\sigma$ is true is not a proof in PA, but rather in a theory, such as ZFC, that is able to theorize about models of PA. So another way to view the theorem is as the claim that if ZFC can prove that a given $\Pi_1$ statement is independent of PA, then ZFC can also prove that it is true.