Let $a,b\in\mathbb R$, $u_0\in C^1(\mathbb R)$ with compact support and $u(t,x):=u(t,x)=e^{at}u_0(bt+x)$ for $t>0$ and $x\in\mathbb R$.
How can we determien the long-time behavior of $u(\;\cdot\;,x)$ for fixed $x\in\mathbb R$?
We clearly need to consider different cases for $a,b$.
Since $\operatorname{supp}u_0$ is bounded, it should hold \begin{equation}\lim_{t\to\infty}u_0(bt+x)=\begin{cases}0&\text{, if }b\ne 0\\ u_0(x)&\text{, if }b=0.\end{cases}\tag{1}\end{equation}
On the other hand, \begin{equation}\lim_{t\to\infty}e^{at}=\begin{cases}0&\text{, if }a<0\\1&\text{, if }a=0\\\infty&\text{, if }a>0.\end{cases}\tag2\end{equation}
How do we determine $\lim_{t\to\infty}u(t,x)$ in the nontrivial case $\alpha\ne0$?
Case $b\neq0$:
after some finite (!) time $T$ (depending only on $x$, $b$ and the support of $u_0$) you will move out of the compact support of $u_0$, hence $u(x,t)=e^{\alpha t}\cdot 0=0$ for $t>T$. In this case the fnite time $T$ trumps the exponential function which tends to infinity only for $t\to\infty$.
Case $b=0$:
exactly as you determined:\begin{equation}\lim_{t\to\infty}u(x,t)=\begin{cases}0&\text{, if }a<0\\u_0(x)&\text{, if }a=0\\u_0(x)\cdot\infty&\text{, if }a>0.\end{cases}\tag2\end{equation} where the last option will evaluate to $0$ if $u_0(x)=0$ and otherwise yield the sign.