Laplace transform of $1/t$

No, it doesn't exist. In general the Laplace transform of $t^n$ is $\frac{\Gamma(n+1)}{s^{n+1}}$, and $\Gamma(n)$ isn't defined on $0,-1,-2,-3...$ This integral is the definition of the Laplace transform, so the transform doesn't exist if the integral doesn't. While there are other integral transforms that could transform $\frac{1}{t}$ in a useful way, anything other than what you gave wouldn't be considered a Laplace transform anymore.

You can actually simplify it further by substituting $st = x$, so you'll get

$$L \left(\frac 1 t \right) = \int \limits _0 ^\infty \frac {\Bbb e ^{-x}} x \ \Bbb d x$$

which is a divergent integral. In other words, the transform doesn't converge for any value of $s$.