We do have $Y_n\sim\frac1a\,\ln n$ in probability as $n\to\infty$.

Indeed, $G(x):=P(X>x)=(c+o(1))e^{-ax}$ as $x\to\infty$, where $c:=Z/a$. So, for any real $p>0$ and $x_n:=\tfrac pa\,\ln n$, $$P(Y_n>x_n)=1-(1-G(x_n))^n=1-(1-(c+o(1))n^{-p})^n \to \begin{cases} 0&\text{ if }p>1 \\ 1&\text{ if }p<1 \end{cases}$$ as $n\to\infty$. $\quad\Box$

We do have $Y_n\sim x_n:=\frac1a\,\ln n$ in probability as $n\to\infty$.

Indeed, $G(x):=P(X>x)=(c+o(1))e^{-ax}$ as $x\to\infty$, where $c:=Z/a$. So, for any real $p>0$, $$P(Y_n>px_n)=1-(1-G(px_n))^n=1-(1-(c+o(1))n^{-p})^n \to \begin{cases} 0&\text{ if }p>1 \\ 1&\text{ if }p<1 \end{cases}$$ as $n\to\infty$. So, $Y_n/x_n\to1$ in distribution and hence in probability. $\quad\Box$