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 $n\to\infty$. $\quad\Box$