Why is the argument of the path integral a pure phase?

That the Boltzmann factor $\exp\left(\frac{i}{\hbar}S\right)$ is a pure phase follows essentially from the fact that the time-evolution operator $U=\exp\left(-\frac{i}{\hbar}H\right)$ is unitary via the general derivation of the path integral formalism from the operator formalism. See also this related Phys.SE post.

Comments to update:

1. There seems to be no reason why OP's eq. (1) should hold.

2. For OP's potential $V=E_0{\rm Ln}\left(1+\left(\frac{x}{x_0}\right)^2\right)$, the Mathematica calculation seems to be calculating the momentum Fourier transform of $\exp\left(\frac{i}{\hbar}V\right)$. There seems to be no reason why that should be a pure phase.