 Solution

For the printed spacetime, yes. The surfaces $T=\text{constant}$, equivalently $t=\text{constant}$, are Cauchy surfaces in Minkowski spacetime. Every inextendible causal curve intersects each once; if its monotone time coordinate had a finite endpoint, the bound $|dx/dT|\le 1$ would give a finite spatial endpoint too, allowing continuation. Causal diamonds are closed and bounded in $(T,x)$ and hence compact. Therefore the printed spacetime is globally hyperbolic.

Initial data on a Cauchy slice determine unique causal evolution for the usual well-posed hyperbolic field equations, without boundary data at spatial infinity at finite time. The time reparametrization preserves that property.

For the qualified AdS interpretation, no. On the universal cover, choose two central events with enough time separation for a null ray to approach a strip edge and return. Their causal diamond contains a sequence with $\rho \to \pi /2$ but no limit in the physical spacetime, so it is noncompact. A constant-$t$ surface fails to meet every inextendible causal curve arriving from timelike conformal infinity. Thus even the cover is not globally hyperbolic. Field evolution needs appropriate boundary conditions, such as a reflecting condition compatible with the field theory, as well as initial data. If AdS time itself is periodically identified, closed timelike curves give an additional and more immediate obstruction. These are consequences of the alternative spatial-lapse geometry, not of the printed flat metric tensor.