Is the Laplace transform essentially a generalized version of the Fourier transform?

A Google request like "differences Fourier Laplace" will provide you many many answers.

I react, very politely, to what @user1952009 has said. For me, and I am not alone, "Fourier Transform" and "Laplace Transform" are very different not only on a mathematical point of view, but also for practical needs.

Fourier Transform belongs to the huge "realm" of harmonic analysis, with a far-reaching extension in group theory.

Whereas Laplace Transform belongs to the domain of complex function theory with abscissas of convergence., etc., that have no equivalent in Fourier Transform theory and vice versa. See (materia.fisica.unipd.it/salasnich/dfl/dfl.pdf)

What is misleading is the similarity of computation rules. But, even on an engineering point of view, one should not say "Fourier = Laplace with $s$ replaced by $2 i \pi \nu$ (or $ i t$)". There are too many differences. One big difference among others: with Laplace, convolutions become products, but not in the other way, whereas, for Fourier, it works in both directions.

The Laplace transform is an important tool to study linear time evolution problems where the situation at time $t=0$ is given. Thus consider ($i$ is added for later convenience) \begin{equation*} \partial _{t}f(t)=iAf(t),\;t\geqslant 0 \end{equation*} where $A$ is some operator. Then the complex Laplace transform \begin{equation*} \hat{f}(z)=\int_{0}^{\infty }dt\exp [izt]f(t),\;Imz>0 \end{equation*} satisfies \begin{equation*} i[z-A]\hat{f}(z)=f(0)\Rightarrow \hat{f}(z)=-i[z-A]^{-1}f(0) \end{equation*} In particular the situation where $A$ is a self-adjoint operator in a Hilbert space is important (think of the Schrödinger equation of quantum mechanics where $A=H$, the Hamiltonian) . The point is that it is much easier to study the properties of $ A$ through its resolvent $[z-A]^{-1}$ than through the time evolution operator $\exp [iAt]$. Actually the Laplace transform above can be considered as a Fourier transform. Thus set \begin{equation*} z=\omega +i\delta ,\;\delta >0 \end{equation*} Then ($\theta (t)$ is the Heaviside step function) \begin{equation*} \hat{f}(\omega +i\delta )=\int_{-\infty }^{+\infty }dt\exp [i\omega t]\theta (t)\exp [-\delta t]f(t) \end{equation*} so we are dealing with the Fourier transform of $\theta (t)\exp [-\delta t]f(t)$. In case $f(t)$ is square integrable this immediately gives the inverse Laplace transform \begin{eqnarray*} \theta (t)\exp [-\delta t]f(t) &=&\frac{1}{2\pi }\int_{-\infty }^{+\infty }d\omega \exp [-i\omega t]\hat{f}(\omega +i\delta ) \\ f(t) &=&\frac{1}{2\pi }\int_{-\infty }^{+\infty }d\omega \exp [-i(\omega +i\delta )t]\hat{f}(\omega +i\delta ) \\ &=&\frac{1}{2\pi }\int_{\Gamma }dz\exp [-izt]\hat{f}(z),\;t\geqslant 0 \end{eqnarray*} where $\Gamma $ is the familiar Bromwich contour, a straight line parallel to and above the real axis.