The answer to your last question is definitely 'no'. The point hotpaw2 makes in his answer is very relevant: the FFT is an efficient implementation of the DFT, and there are no equivalently efficient implementations for the numerical computation of the $\mathcal{Z}$-transform or the Laplace transform.
But that's not the only reason. There are important functions (or sequences) for which the Laplace transform or the $\mathcal{Z}$-transform don't even exist, whereas the Fourier transform does. E.g., take a sinusoid or a complex exponential extending from $-\infty$ to $\infty$. These functions (or sequences) don't have a Laplace transform or a $\mathcal{Z}$-transform. Other important examples are impulse responses of ideal frequency selective filters such as low pass or high pass. They can be represented in terms of the sinc function, which can only be transformed using the Fourier transform, but not using the (bilateral) Laplace transform or - in discrete time - the (bilateral) $\mathcal{Z}$-transform.
So even if formally it looks like the Fourier transform is a special case of the Laplace transform or the $\mathcal{Z}$-transform, that's generally not case. One reason for that is the incorporation of the theory of distributions in the theory of the Fourier transform (i.e., the use the Dirac delta impulse), which makes it possible to compute the transform of functions like $\sin(\omega_0t)$ or $e^{j\omega t}$. The latter is not possible using the (bilateral) Laplace transform (or, in discrete time, the bilateral $\mathcal{Z}$-transform).
When people see the definitions of the (bilateral) Laplace transform and of the Fourier transform
$$X_L(s)=\int_{-\infty}^{\infty}x(t)e^{-st}dt\\
X_F(j\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt\tag{1}$$
it may seem obvious to them that both transforms become identical by substituting $s=j\omega$. This, however, is generally not true. The pitfall here is the fact that the substitution does not take into account the convergence of the improper integrals. Depending on $x(t)$, the Laplace integral might not converge for $s=j\omega$, so $X_F(j\omega)$ might not even exist. The substitution is only valid if the region of convergence (ROC) of the Laplace integral includes the $j\omega$-axis. A completely analogous argument is true for the $\mathcal{Z}$-transform and the DTFT. In that case the substitution $z=e^{j\omega}$ is only valid if the ROC includes the unit circle.
The last paragraph may seem to imply that the Laplace transform and the $\mathcal{Z}$-transform are simply more general than the respective versions of the Fourier transform. However, this is also not true, as already mentioned above, because there are functions (sequences) that can only be treated by the Fourier transform, but not by the Laplace transform ($\mathcal{Z}$-transform).
Also take a look at the following answers to related questions: answer 1, answer 2, answer 3.