There's a couple approaches to computing the Laplace transform of $e^t$ but the "slickest" way is just to use the derivative rule: $\mathcal{L}_t[f'(t)](s) = s\mathcal{L}_t[f(t)](s) - f(0).$ Then you can notice that the derivative of $e^t$ is just $e^t$, so you can solve for $\mathcal{L}_t[e^t]$ in the above equation. This gives $\mathcal{L}_t[e^t] = \frac{1}{s-1}$.
The derivative rule is just computed with integration by parts, which you could use for $e^t$ directly. See this page.
If you look at the page I linked, it sates:
The Laplace transform existence theorem states that, if $f(t)$ is piecewise continuous on every finite interval in $[0,\infty)$ satisfying $f(t) \leq Me^{at}$ for all $t\in[0,\infty)$, then $\mathcal{L}_t[f(t)](s)$ exists for all $s > a$.
So basically, given $a$ you need to find an $M$ such that $\log M \geq (1-a)t$ for every $t\in[0,\infty)$. You can obviously do this when $a \geq 1$ because the RHS is always negative or $0$ in that case. If $a < 1$ you can't, because the RHS is unbounded. So $a = 1$ is the best choice here and $\mathcal{L}_t[f(t)](s)$ exists for $s > 1$.