Is the Hilbert transform computable "losslessly"?
The loss you are seeing is due to the use of a finite length rectangular window.  If your signal is long enough that you can ignore the Hilbert filter transients at the both ends of the window (e.g. if it is infinite in length or so long that you don't care about the ends below some noise floor), then the process won't have the particular visible losses that your example points out.