Laplace transforms may be considered to be a super-set for CTFT (Continuous-Time Fourier Transforms). You see, on a ROC (Region of Convergence) if the roots of the transfer function lie on the imaginary axis, i.e. for s=σ+jω, σ = 0, as mentioned in previous comments, the problem of Laplace transforms gets reduced to Continuous Time Fourier Transform. To rewind back a little, it would be good to know why Laplace transforms evolved in the first place when we had Fourier Transforms.
You see, convergence of the function (signal) is a compulsory condition for a Fourier Transform to exist (absolutely summable), but there are also signals in the physical world where it is not possible to have such convergent signals. But, since analysing them is necessary, we make them converge, by multiplying a monotonously decreasing exponential e^σ to it, which makes them converge by its very nature. This new σ+jω is given a new name 's', which we often substitute as 'jω' for sinusoidal signals response of causal LTI (Linear Time-Invariant) systems.
In the s-plane, if the ROC of a Laplace transform covers the imaginary axis, then it's Fourier Transform will always exist, since the signal will converge. It is these signals on the imaginary axis which comprise of periodic signals e^jω = cos ωt + j sin ωt (By Euler's).
Much in the same way, z-transform is an extension to DTFT (Discrete-Time Fourier Transforms) to, first, make them converge, second, to make our lives a lot easier. It's easy to deal with a z than with a e^jω (setting r, radius of circle ROC as untiy).
Also, you are more likely to use a Fourier Transform than Laplace for signals which are non-causal, because Laplace transforms make lives much easier when used as Unilateral (One sided) transforms. You could use them on both sides too, the result will work out to be the same with some mathematical variation.