It depends on what you mean by "the system" being invariant here. If you mean the Hamiltonian with a specific choice of $A$, then that's not translation invariant. However, the equations of motion do not depend on the choice of $A$, and they are translation-invariant in both directions. You can argue that, classically, the equations of motion are what matters, but when you quantize you have no choice but to choose a Hamiltonian.
Symmetries of the equation of motion need not necessarily lift to the Hamiltonian, as we can plainly see here and as is discussed more generally here by Qmechanic. One can show that there is no way to make both canonical momenta conserved for any choice of $A$, see this answer of mine.
This particular situation is also discussed at length and excellently by Emilio Pisanty in this answer. Summarizing, this Hamiltonian system (regardless of the choice of $A$), has two conserved quantities, namely the $(x,y)$-coordinate of the center of the circle the moving particle describes. As usual in Hamiltonian mechanics, conserved quantitites generate symmetries. The choice of $A$ is unphysical and hence cannot change these quantities, but what the choice of $A$ does is "rotate" the canonical momenta $p_x,p_y$. So the Hamiltonian is translation invariant in one direction exactly when one of the canonical momenta aligns with one of the conserved quantities, but, in general, the transformation generated by them will not be translation.
This underscores that the canonical momentum of Hamiltonian mechanics is, in general, a purely fictitious quantity. There is no general physical interpretation for it and transformations that do not change the physics, such as the choice of $A$, may change it from a physically meaningful conserved quantity to...just something else that describes the a point in phase space.