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. 
$\quad\Box$