Do the eccentricity of a binary orbit change with change in the semimajor axis due to loss of orbital angular momentum by gravitational radiation?
Yes. There's an intuitive way to see this without reproducing all of the messy GR equations (which appear in a Physics StackExchange answer pointed to by @PM2Ring in a comment).
The amount of gravitational radiation (or graviational waves, GW) emitted is proportional to the instantaneous acceleration of an object (well second derivative of the mass quadrupole moment...). Imagine a highly eccentric orbit, like Haley's comet (link to .gif). At the closest approach of the orbit, periapsis, the orbit has maximum speed and undergoes a sharp turn. This is the point of maximal acceleration. At apoapsis the orbit is very slow, so even with the turn the acceleration is minimal.
At periapsis there will be a big burst of GWs. Everywhere else when the acceleration is small there will be minimal GW emission.
The burst of GWs at periapsis reduces the total energy of the system. This means the next apoapsis will be closer to the center of mass. The orbiting object can't get as far away with less energy. After reaching the new apoapsis, it falls back and emits another burst at periapsis. Over time the distance of the apoapsis will approach the distance of the periapsis, circularizing the orbit.
Even for a near circular orbit, an asymmetry in GW emission remains, bringing the orbit ever closer to a circle.
For an object like Haley's comet, the periapsis speed isn't fast enough to emit noticeable gravitational radiation, so GW emission doesn't really affect the orbit. But this is why we expect most binary black hole (BBH) or neutron star systems in the LIGO band to have relatively small eccentricity. By the time the binary approaches merger the orbits have pretty well circularized. Fig 1 from this paper by Zevin, et al (2021) shows that very few BBHs will have eccentricity more than $0.1$ by the time their GW emission reaches $10$ Hz.