The Laplace transform of a function $f(t)$ is a function that maps $\mathbb{C} \mapsto \mathbb{C}$.
$$f(s) = \int_0^\infty f(t)e^{-st}dt, \text{ with } s=x + iy$$
Since $s = x + iy$ is complex, $e^{-s}$ can be written as $e^{-x}e^{-iy}$. Am I correct that this can be interpreted as an amplitude multiplied with an angle represented by the unit vector $e^{-iy}$? Or is it amplitude multiplied by frequency?
What should the result of $f(s)$ be interpreted as?
You interpretation of $e^{-s}$ is pretty much correct, at least if interpreted somewhat liberally. Now, if you fix $x$, we are looking at the Fourier transform of the function $t\mapsto f(t)e^{-xt}$. In that sense, $y$ is rightly interpreted as a frequency. So if you think you understand the Fourier transform, that gives you an idea what the Laplace transform means. However, the extra factor $e^{-xt}$ complicates things.
In practice, it doesn't seem fruitful to hunt for an interpretation of the Laplace transform. It is just a transform which changes operations like integration, differentiation, and convolution into simpler algebraic operations, and doesn't have a lot of meaning beyond that.