Ok, here's my (not very rigorous) try. Other approaches or refinements are welcomed.
Let's change notation $m = a n$, $x = n = mb$ with $b = 1/a$.
Then $$e^x = \sum_{k=0}^m \frac{x^k}{k!} + R_m(x) \tag1$$
with the remainder:
$$
\begin{align}
R_m(x) &= \int_0^x \frac{(x-t)^m}{m!} e^t dt\\
&=\frac{e^x}{m!}\int_0^x y^m e^{-y} dt\\
&=\frac{e^x}{m!} \left(m!- \int_x^\infty y^m e^{-y} dt\right) \tag2\\
\end{align}
$$
Because $b>1$, we can approximate the complement of the gamma integral by abusing Laplace's method. Namely, for a differentiable positive decreasing function (more in general, a function that has its global maximum at the start of the interval of integration) and for $m\to \infty$ we approximate
$$ \int_c^\infty e^{m h(x)}dx\approx \int_c^\infty e^{m [h(c) + h'(c)(x-c)]}dx=\frac{e^{m \, h(c)}}{m \,|h'(c)|} \tag{3}$$
Then we can write
$$
\int_x^\infty y^m e^{-y} dt =\int_x^\infty e^{m (\log(y)-y/m)} \approx x^m e^{-x} \frac{b}{b-1} \tag 4\\
$$
Actually we are abusing the method here because our $h()$ depends also on $m$ - this would need some justification. Passing over this and putting all together:
$$\begin{align}
\sum_{k=0}^m \frac{x^k}{k!} &= e^x - R_m(x) \\
& \approx \frac{x^m}{m!} \frac{b}{b-1} \\ \tag{5}
&= \frac{n^{an}}{(an)!} \frac{1}{1-a} \\
& \approx \left(\frac{e}{a}\right)^{an} \frac{1} {(1-a) \, \sqrt{ 2 \pi a} \, \sqrt{n}} \tag{6}
\end{align}
$$
Finally
$$\lim_{n\to \infty} \left(\sum\limits_{k=0}^{an} \frac{n^k}{k!} \right)^{1/n}= \left(\frac{e}{a}\right)^a \tag 7$$
Added: As rightly comments @Maxim, if we are interested in correcting the rounding (when $m$ in $(5)$ is not an integer we round down to the nearest integer), we should multiply $(6)$ by the correction factor $a^{\{an\}}$,
where $\{\}$ denotes the fraction part. Of course, this correction is asymptotically negligible ($O(1)$) and does not change the limit $(7)$.