Edward Green wrote:
Suppose we had two masses in relative motion, so that at least one
mass must be in motion in any coordinate system (at least in
approximately locally Lorentz coordinates which encompass both
masses).
At least one of the masses therefore has kinetic energy. Must we
include this kinetic energy in the stress energy tensor, or can we
simply use the rest mass of each as the energy, so long as we treat
the masses discretely?

The energy-momentum tensor automatically includes the motion of an
object. One must include both its kinetic energy and its momentum, in
addition to its mass. In relativity, a pointlike mass m contributes to
the energy-momentum tensor:
T^ij(x) = m delta^4(x-X(t))) U^i U^j
Where U is the 4-momentum of the object and X(t) is its trajectory; x
represents the 4 coordinates on spacetime.

When projected onto the locally inertial frame in which it is at rest,
only U^00 is nonzero, and corresponds to its mass density.


Follow on question: assuming the answer is that we may treat each
mass as contributing its rest mass alone, [...]

You cannot do so, except in its rest frame.


Eventually our independent masses will appear to be a swarm of dust
particles, and we will elect to treat then on average, rather than
discretely. [...]

Yes. A collection of non-interacting pointlike masses is called "dust",
and the energy-momentum tensor for dust with a mass density \rho is:
T^ij(x) = \rho(x) U^i U^j
Where U is the 4-momentum of the infinitesimal element at x.


Since both pressure and kinetic energy stem from the same root: that
the particles are in relative motion, are we over-counting if we
include both?

Not if you do it correctly.


Tom Roberts