That depends on how you define 'conservation of energy'. Consider an arbitrary bounded volume of spacetime. Conservation of energy says that the net flow across the boundary of any such volume is 0. Under this definition moving in space but not time is not a violation, as the same amount of mass enters the volume as exits it. The only odity is that both events happen at the same time
The idea is that the volume we are talking about is a 4 dimensional volume of space-time and is bounded; not a three dimensional volume of space that extends to infinity along time.
Conservation of energy says that it is impossible for energy to enter this volume with that same amount of energy exiting the volume.
Consider what it would mean for this to be violated. For the sake of argument, assume that all particles must move forward in time by a non zero amount at all points along there path. Since the volume is bounded, any particle with an infinite path must eventually have a time coordinate beyond the largest time coordinated in the volume. Therefore, the particle must eventually exit the volume. If you were to work out the geometry more carefully, you could show with relative ease that the particle must exit the volume an equal number of times as it enters. If a particle were to enter the volume without exiting the volume, it would mean that said particle was destroyed within the volume. Similarly, if a particle were to exit the volume without entering, it would have to have been created within the volume. Both of these situations are possible if an interaction occurs within the volume, but the net energy of the particles leaving such an interaction, must be the same as the net energy of the particles entering the interaction.
Put another way, assume that all interactions obey the conservation of energy. If our original volume was V, we can construct a new volume V' from V by carving out sub volumes in which an interaction occurs. Since all such sub volumes obey the conservation of energy (by assumption), the net energy flow into and out of V' must be the same as for V. However, since no interactions occur withing V', all particles entering V' must exit V' an equal number of times, so the net energy flow of V' must be 0. Therefore the net flow of V must also be 0.
