For a complex variable $s=a+ib$ and the condition $\Re\{s\}<0$ we have the limit $$\lim_{t\to\infty}e^{st}\to 0$$ Question: Why no condition for the imaginary part?
If $\Re\{s\}>0$ we have $a>0$, and if also $\Im\{s\}<0$ we have $b<0$, so why isn't $$ \lim_{t\to \infty}e^{(a+ib)t} =\lim_{t\to \infty}e^{at}e^{ibt} \to"\infty\cdot 0" \to 0\quad ? $$
Because $$|e^z|=|e^{a+ib}|=|e^a||e^{ib}|=|e^a|$$ therefore practically the real part of $z$ is useful and effective here. With your assumption $a>0$ $$|e^{(a+ib)t}|=|e^{at}|\to\infty$$ as $t\to\infty$.