Electrical – Convert Current source to Voltage source in Laplace domain

The Fourier and the Laplace transform are not the same. First of all, note that when we talk about the Laplace transform, we very often mean the unilateral Laplace transform, where the transformation integrals starts at \$t=0\$ (and not at \$t=-\infty\$), i.e. with the Laplace transform we usually analyze causal signals and systems. With the Fourier transform this is not always the case.

In order to understand the differences between the two, it is important to look at the region of convergence (ROC) of the Laplace transform. For causal signals, the ROC is always a right-half plane, i.e. there are no poles (of a rational function in \$s\$) to the right of some value \$\sigma_0\$ (where \$\sigma\$ denotes the real part of the complex variable \$s\$). Now if \$\sigma_0<0\$, i.e. if the \$j\omega\$ axis is inside the ROC, then you simply obtain the Fourier transform by setting \$s=j\omega\$. If \$\sigma_0>0\$ then the Fourier transform does not exist (because the corresponding system is unstable). The third case (\$\sigma_0=0\$) is interesting because here the Fourier transform does exist but it cannot be obtained from the Laplace transform by setting \$s=j\omega\$. Your example is of this type. The Laplace transform of the step function has a pole at \$s=0\$, which lies on the \$j\omega\$ axis. In all such cases the Fourier transform has additional \$\delta\$ impulses at the pole locations on the \$j\omega\$ axis.

Note that it is not true that the Fourier transform cannot deal with transients. This is just a misunderstanding which probably comes from the fact that we often use the Fourier transform to analyze the steady-state behavior of systems by applying sinusoidal input signals that are defined for \$-\infty<t<\infty\$. Please also see this answer to a similar question.

Elaborating on Andy aka's answer ...

You need the transform pair

$$\frac{\omega_0}{(s+\alpha)^2+\omega_0^2}\Longleftrightarrow e^{-\alpha t}\sin(\omega_0t)u(t)$$

with

$$\omega_0^2=\frac{1}{LC}-\frac{R^2}{4L^2}$$

and

$$\alpha=\frac{R}{2L}$$

Of course you'll also get some multiplicative constant, but that's straightforward.