Why the Fourier and Laplace transforms of the Heaviside (unit) step function do not match?

Actually they do match in the sense that the Laplace transform provides an analytic continuation of the Fourier transform result to the complex plane. Look at the limits of the real and imaginary parts of

$\frac{1}{s}=\frac{s^{*}}{|s|^2}=\frac{\sigma-i\omega}{\sigma^2+\omega^2}$

as the real part of $s$ tends to $0$. There's no discrepancy; you are looking at a one-dimensional slice of a two-dimensional function (the blind men and the elephant allegory).

Hints: Look at MSE-122220 and MSE-73922. Also think of the Cauchy contour integral of $f(z)/z$ with the contour being a rectangle about the origin that gradually and symmetrically extends to infinity in length and collapses in height to the real line. Additional info available at Wiki on the Cauchy principle value and the Poisson kernel rep of the nascent delta function.

PS: The bilateral Laplace transform equals the unilateral Laplace transform when acting on H(t)f(t) where H(t) is the Heaviside step function. In this case, letting $s= \sigma+i\omega$ clearly shows that the Laplace transform provides an analytic continuation in general of the FT result to the complex plane for $\sigma>0$.

The integral defining $\mathcal{L}(x)$ converges for all $s>0$ (or, more generally, for all $s\in\mathbb{C}$ with $\operatorname{Re}(s)>0$.) However, the integral defining $\mathcal{F}(x)$ does not converge for any $\omega\in\mathbb{R}$. As noted n the comments, $\mathcal{F}(x)$ is not a function, but a distribution; it is defined not as an integral, but through a different process (duality.)

Another way of seeing this, is that $\mathcal{L}(x)(i\,\omega)$ is not defined for any $\omega\in\mathbb{R}$.