Under what conditions does jw equal the laplace variable s in an electrical circuit?
The Laplace variable \$s\$ relates to Fourier's \$j\omega\$ as follows:

$$
s = \sigma + j\omega
$$

Fourier transform can be seen as a Laplace transform when \$\sigma=0\$. The \$\sigma\$ allows the Laplace integral transformation to converge for signals that Fourier transform does not, e.g. a unitary step (Heaviside function).

If you are working with real signals, in a steady-state regime, chances are that waveforms will converge for both Fourier and Laplace (the signal won't suddenly become unbounded or will present a non-differentiable spot), so \$s\$ and \$j\omega\$ become interchangeable. In a mathematically rigorous fashion, the Laplace ROC (Region of Convergence) must include \$\sigma=0\$, also known as the \$j\omega\$ axis of the s-plane.

The convergence limitations of Fourier are grounds for some hardcore maths, if you are interested, Wikipedia has a great link on this.