We must be careful when expanding in small $a$, we do not know a priori that $a x$ is small. It is in fact necessary to treat $a$ as small, and $a x$ to be at least of order one. While the maximum value of the integral goes like $\ln a^{-1}$, there is an error of order one in the accepted answer.
Change variables. Let $z = a y$. The integral can then be written as $I(a,x) = (a x)^{-a}\left[\Gamma(a,a) - \Gamma(a,a x)\right],$ where $\Gamma(s,x) = \int_x^\infty dt \ t^{s-1} e^{-t}$ is the incomplete gamma function.
Look for an extremum, $X$. (The reader can verify the extremum found below is a maximum.) We find ${\partial I}/{\partial x}|_{x=X} = 0$ implies \begin{equation*} a\Gamma(a,a X) + (a X)^a e^{-a X} = a \Gamma(a,a) \tag{1} \end{equation*} so that \begin{equation*} I(a,X) = \frac{1}{a} e^{-a X}. \end{equation*} Integrating $a\Gamma(a,a X)$ in (1) by parts we find the condition on $X$ is $\frac{1}{a}\Gamma(a+1,a X) = \Gamma(a,a).$ So far no expansion has been made. Expanding the left hand side in small $a$ (but not small $a x$), we find $\begin{eqnarray} \frac{1}{a}\Gamma(a+1,a X) &=& \frac{1}{a}e^{-a X} + \mathrm{h.o.} \\ &=& I(a,X) + \mathrm{h.o.}. \end{eqnarray}$ We find the higher order terms go like $e^{-a X} \ln a X$. We assume $a X$ is large enough so these terms are suppressed. We will find this assumption to be self consistent. Therefore, $I(a,X) = \Gamma(a,a) + \mathrm{h.o.}$. Expanding $\Gamma(a,a)$ in small $a$ we find $\begin{eqnarray} I(a,X) &=& -\mathrm{Ei}(-a) + \mathrm{h.o.} \\ &=& -\gamma + \ln a^{-1} + \mathrm{h.o.}, \end{eqnarray}$ where $\mathrm{Ei}(x) = \int_{-\infty}^x d t \ t^{-1} e^t$ is the exponential integral and $\gamma$ is the Euler-Mascheroni constant. The location of the maximum is $X = -\frac{1}{a} \ln\left[a(\ln a^{-1}-\gamma)\right] + \mathrm{h.o.}$ For $a = e^{-k}$ we find $a X \approx k$ so the higher order terms discussed above go like $e^{-k}\ln k$. Thus, since $k$ is large the expansion is valid.
Below is a plot of $I(a,x)$ for $a=10^{-4}$. The predicted maximum is $I(a,X) \approx 8.63$ at $X \approx 7.05\times 10^4$. Clearly $\ln a^{-1} -1 \approx 8.21$ underestimates the maximum value of the integral.
