Digital filters and the Z-transform
The S-transform allows you to deal with differential equations in an algebraic manner - so they become easier to solve. Since continuous/analog filters consist of integrators and differentiators the S-transform is therefore a natural way to deal with these systems.

The z-transform provides an algebraic way of dealing with finite difference systems and therefore it is a natural way to deal to discrete-time systems i.e. digital.

In the S-transform, setting $s=j\omega$ results in the Fourier transform. For continuous time systems we are interested in whether poles are on the right or left side of the $j\omega$ axis, because that determines the stability.

For the Z-transform, setting $z=e^{j\omega}$ results in the Discrete-Time Fourier transform. To determine stability, we are interested in whether the poles are inside or outside the unit circle.

So the S and Z domains are similar - they allows you solve continuous and discrete time systems, respectively,  using algebra.