Since shock waves have non-negligible viscosity, why do the normal shock relations ignore friction/viscosity?
Neglecting friction does indeed lead to inaccuracies. Those, however, are small for weak shocks, so using the equations for inviscid shocks gives a good approximation. Especially in the times before electronic computers, this was a major simplification while the error was tolerably small.

Stronger shocks should be avoided and replaced by a cascade of weaker, oblique shocks in order to minimize losses, so for practical work there is little need for equations which include those losses. Only in academic work and the design of reentry vehicles with their blunt geometries would the precise calculation of stronger shocks be of any interest.

But how large are those losses? Wikipedia has a page on normal shock tables and the last column gives the ratio of the stagnation pressures ahead and past the shock. Since both should be the same in case of a lossless shock, the real number gives a good approximation of the energy losses incurred over the shock. The more the result deviates from 1, the bigger the losses. 1% of stagnation pressure is lost at a Mach number ahead of the shock of approximately 1.225. 10% is lost at Mach 1.587. It goes down from there: 50% is lost at Mach 2.5 and 90% at Mach 4.4. All values use a ratio of specific heats of 1.4, so they are valid for a two-atomic gas (or a mixture of predominantly two-atomic gasses like air).