No.

Indeed, let $\mathcal{V}_n$ be the variety of magmas generated by the relators with variable $y$ saying that for every $k\le n$, all products of $k$ copies of $y$ are equal. Since the variety of power-associative magmas is $\bigcap_n \mathcal{V}_n$, a negative answer to the question is equivalent to showing that for every $n$ the relatively free magma $M_n$ on 1-generator $x$ in $\mathcal{V}_n$ is not associative.

Write by induction $x^1=x$, $x^k=xx^{k-1}$. I claim that in $M_n$ we have $x^{n+1}\neq x^nx$. Indeed, $x^nx$ can be rewritten in all ways $(ab)x$ with $a,b$ products of $k,\ell$ copies of $x$ (in some order) for $k+\ell=n$. No such $(ab)x$ is a relator, and no nontrivial relator has the form $(x=\dots)$. Hence $(ab)x$ can only be transformed by a relator substitution into another $(a'b')x$.

(This actually shows that $\mathcal{V}_{n+1}$ is properly contained in $\mathcal{V}_n$.)

No.

Indeed, let $\mathcal{V}_n$ be the variety of magmas generated by the relating identities with variable $y$ saying that for every $k\le n$, all products of $k$ copies of $y$ are equal. Since the variety of power-associative magmas is $\bigcap_n \mathcal{V}_n$, a negative answer to the question is equivalent to showing that for every $n$ the relatively free magma $M_n$ on 1-generator $x$ in $\mathcal{V}_n$ is not associative.

Write by induction $x^1=x$, $x^k=xx^{k-1}$. I claim that in $M_n$ we have $x^{n+1}\neq x^nx$. Indeed, $x^nx$ can be rewritten in all ways $(ab)x$ with $a,b$ products of $k,\ell$ copies of $x$ (in some order) for $k+\ell=n$. No such $(ab)x$ is a relator, and no nontrivial relator has the form $(x=\dots)$. Hence $(ab)x$ can only be transformed by a relator substitution into another $(a'b')x$.

(This actually shows that $\mathcal{V}_{n+1}$ is properly contained in $\mathcal{V}_n$.)