The Laplace transform is designed to deal with causal problems in which some input is provided to a linear system at $t=0$, and the system provides a response for $t \gt 0$. There is no response for $t \lt 0$, as the phenomenon being modeled is causal. Thus, when we invert the Laplace transform, we adjust the real part of the line along which we integrate as an expression of this causality.
But why to the right of the rightmost pole? Consider a LT $F(s)$ which has isolated poles in the complex plane and no other singularities. We can compute the ILT using the Residue Theorem by either closing to the left or to the right. If we close to the left, then the contour integral
$$\oint_C dz F(z) e^{z t} = \int_{\sigma-i R}^{\sigma+i R} ds F(s) e^{s t} + i R \int_{\pi/2}^{3 \pi/2} d\theta \, e^{i \theta} F \left ( R e^{i \theta} \right ) e^{R t \cos{\theta}} e^{i R t \sin{\theta}} $$
(NB there is a pair of segments from $\sigma \pm iR$ to the circle $z=R e^{i \theta}$, but one can easily show that the integral over these segments vanishes as $R \to \infty$.)
We want the second integral to vanish as $R \to \infty$ so that we may express the ILT in terms of the residues of the poles of $F$. Note that, when we close to the left, $\cos{\theta} \lt 0$, so that the second integral may vanish only when $t \gt 0$. ($F$ must also satisfy certain conditions as $R \to \infty$, but $t$ must be greater than zero in any case.)
Note that when $t \lt 0$, we may not close the contour to the left. Rather, we must close to the right in order to use the Residue Theorem. Over that semicircle, $\cos{\theta} \gt 0$, so the second integral vanishes only when $t \lt 0$.
Thus, to capture causality, we move the line over which we define the integral for the ILT such that all poles are to the left of the line; in that way $f(t) =0$ when $t \lt 0$, as one would expect.
I think you think this question too complicated.
Similar to how to solve this PDE:
Let $\begin{cases}p=x+ct\\q=x-ct\end{cases}$ ,
Then $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial x}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial x}=\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}$
$\dfrac{\partial^2u}{\partial x^2}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}\right)=\dfrac{\partial}{\partial p}\left(\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}\right)\dfrac{\partial p}{\partial x}+\dfrac{\partial}{\partial q}\left(\dfrac{\partial u}{\partial p}+\dfrac{\partial u}{\partial q}\right)\dfrac{\partial q}{\partial x}=\dfrac{\partial^2u}{\partial p^2}+\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}=\dfrac{\partial^2u}{\partial p^2}+2\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}$
$\dfrac{\partial u}{\partial t}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial t}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial t}=c\dfrac{\partial u}{\partial p}-c\dfrac{\partial u}{\partial q}$
$\dfrac{\partial^2u}{\partial t^2}=\dfrac{\partial}{\partial t}\left(c\dfrac{\partial u}{\partial p}-c\dfrac{\partial u}{\partial q}\right)=\dfrac{\partial}{\partial p}\left(c\dfrac{\partial u}{\partial p}-c\dfrac{\partial u}{\partial q}\right)\dfrac{\partial p}{\partial t}+\dfrac{\partial}{\partial q}\left(c\dfrac{\partial u}{\partial p}-c\dfrac{\partial u}{\partial q}\right)\dfrac{\partial q}{\partial t}=c^2\dfrac{\partial^2u}{\partial p^2}-c^2\dfrac{\partial^2u}{\partial pq}-c^2\dfrac{\partial^2u}{\partial pq}+c^2\dfrac{\partial^2u}{\partial q^2}=c^2\dfrac{\partial^2u}{\partial p^2}-2c^2\dfrac{\partial^2u}{\partial pq}+c^2\dfrac{\partial^2u}{\partial q^2}$
$\therefore\dfrac{\partial^2u}{\partial p^2}+2\dfrac{\partial^2u}{\partial pq}+\dfrac{\partial^2u}{\partial q^2}-\dfrac{1}{c^2}\left(c^2\dfrac{\partial^2u}{\partial p^2}-2c^2\dfrac{\partial^2u}{\partial pq}+c^2\dfrac{\partial^2u}{\partial q^2}\right)=\delta\left(\dfrac{p+q}{2}\right)f\left(\dfrac{p-q}{2c}\right)$
$4\dfrac{\partial^2u}{\partial pq}=\delta\left(\dfrac{p+q}{2}\right)f\left(\dfrac{p-q}{2c}\right)$
$\dfrac{\partial^2u}{\partial pq}=\dfrac{1}{4}\delta\left(\dfrac{p+q}{2}\right)f\left(\dfrac{p-q}{2c}\right)$
$u(p,q)=F(p)+G(q)+\dfrac{1}{4}\int_b^q\int_a^p\delta\left(\dfrac{r+s}{2}\right)f\left(\dfrac{r-s}{2c}\right)dr~ds$
$u(x,t)=F(x+ct)+G(x-ct)+\dfrac{1}{4}\int_b^{x-ct}\int_a^{x+ct}\delta\left(\dfrac{r+s}{2}\right)f\left(\dfrac{r-s}{2c}\right)dr~ds$
Best Answer
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 $\sigma>0$.