Photon pair production at relativistic speeds?

Yes, there is a simple explanation! The pair production process $\gamma \to e^+ + e^-$ is forbidden by energy-momentum conservation, so it doesn't happen in either frame.

One way to see this is that, as a massless object, the photon has "the most possible momentum for its energy", as it doesn't have the extra rest mass-energy. So if you try to make the momentum in this equation balance, the electron and positron will have too much energy, and vice versa.

However, a slightly modified version of your idea holds. Our universe is full of low-energy photons from the Cosmic Microwave Background. As you suggested, a very fast moving charged particle will see these photons as heavily blueshifted, and pair production is possible if the photons scatter off the object. (This is a $2 \to 2$ reaction, so the argument in the previous paragraph doesn't hold.)

The scattering steals some of the particle's energy, slightly slowing it down. This places a limit on the energy of cosmic rays, called the GZK bound.

Pair production cannot just happen. It requires some other particle/object to balance out the momentum. Lets suppose this object is a nucleus.

So the interaction looks to the fast moving observer like a blue shifted photon scattering off a stationary nucleus and producing an $e\bar{e}$ pair. To us it looks like a nucleus moving at relativistic speed scattering a photon and producing an $e\bar{e}$ pair. There is no fundamental disagreement between the observers as to what happens.