If you mean "not frame-invariant" in the sense that it's not a scalar, true. But it is certainly "frame-invariant" in the sense that it's a tensor and transforms accordingly when you change coordinates, so contracting it with other tensors yields frame-invariant scalars. Any actual observable that tells you about the "amount of gravity produced" by an object will be such a scalar, so it will be frame-invariant, as I said.
No. The "amount of gravity produced" by the object is not a function of its energy, it's a function of its stress-energy tensor, of which energy is only one component. In a frame in which the object is moving, there will be other non-zero components of the SET as well as the energy, and their effects will offset the apparent "effect" of the increase in energy, so the final result will be the same as it is for a frame in which the object is at rest.
This is true; so far I've only talked about a single object. Since GR is nonlinear, two solutions do not add up to another solution, so I can't just take the solution for each body in isolation and add them to get a solution for both bodies. That's why we don't have a closed form solution for, e.g., binary pulsars, but have to calculate their expected orbital changes due to the radiation of gravitational waves numerically.
However, we don't have to get into that to resolve the question of whether an object being in motion changes the gravity it produces. It doesn't.
This is key: this is almost always the quickest method of figuring out how much gravity an object produces. Transform to its rest frame, in which the stress-energy tensor will usually have its simplest form. Actually, in the case given in the OP, it's even simpler, since each object is isolated so the individual solution for the gravity produced by the object, in the vacuum region outside the object, is just the Schwarzschild solution with the object's mass M. Transforming that solution to a frame in which M is moving does not change any of its physical predictions, it just makes it look more complicated while still giving the same answers.