How is Wilkinson power divider lossy?
The ideal Wilkinson does not have any loss when transferring signals from port 1 to ports 2 and 3:
$$|S_{11}|^2 + |S_{21}|^2 + |S_{31}|^2 = 1$$
When the input is given at port 2 or port 3, it is not lossless:
$$|S_{12}|^2 + |S_{22}|^2 + |S_{32}|^2 \ne 1$$
Therefore we don't classify it as a lossless network in the general sense, even though it is lossless in its usual application.