Is the Laplace transform redundant?

The Laplace transform is a generalization of the Fourier transform since the Fourier transform is the Laplace transform for $s = j\omega$ (i.e. $s$ is a pure imaginary number = zero real part of $s$).

Reminder:

Fourier transform: $X(\omega) = \int x(t) e^{-j\omega t} dt$

Laplace transform: $X(s) = \int x(t) e^{-s t} dt$

Besides, a signal can be exactly reconstructed from its Fourier transform as well as its Laplace transform.

Since only a part of the Laplace transform is needed for the reconstruction (the part for which $\Re(s) = 0$), the rest of the Laplace transform ($\Re(s) \neq 0$) seems to be unuseful for the reconstruction...

Is it true?

Also, can the signal be reconstructed for another part of the Laplace transform (e.g. for $\Re(s)=5$ or $\Im(s)=9$)?

And what happens if we compute a Laplace transform of a signal, then changing only one point of the Laplace transform, and compute the inverse transform: do we come back to the original signal?