Relation between Laplace and Fourier transforms
The Laplace transform evaluated at $s=j\omega$ is equal to the Fourier transform if its region of convergence (ROC) contains the imaginary axis. This is also true for the bilateral (two-sided) Laplace transform, so the function need not be one-sided.

As for real and imaginary parts, since $s$ is a complex variable, both the Laplace and the Fourier transform generally have real and imaginary parts. Take as a simple example the function $x(t)=e^{-at}u(t)$, with $a&gt;0$, where $u(t)$ is the unit step function. The Laplace transform is

$$X_L(s)=\frac{1}{s+a}\tag{1}$$

Since $a&gt;0$, the ROC of $X_L(s)$ contains the imaginary axis, and the Fourier transform of $x(t)$ is simply obtained by evaluating $X_L(s)$ on the imaginary axis $s=j\omega$:

$$X_F(\omega)=X_L(j\omega)=\frac{1}{j\omega+a}\tag{2}$$

Since $s=\sigma+j\omega$ is generally complex, not only the Fourier transform but also the Laplace transform $(1)$ has a real and an imaginary part:

$$X_L(\sigma+j\omega)=\frac{1}{\sigma+j\omega+a}=\frac{\sigma+a}{(\sigma+a)^2+\omega^2}-j\frac{\omega}{(\sigma+a)^2+\omega^2}\tag{3}$$

Only when evaluated on the real axis $s=\sigma$ ($\omega=0$) does the imaginary part vanish.