 Solution

Yes. Removing zero leaves the induced order strict and linear, with no endpoints. For any $a<b$ there are infinitely many points in the real interval between them, so one can choose an intermediate point different from zero. More explicitly, the midpoint works unless it is zero, in which case $(2a+b)/3$ is nonzero and lies strictly between the endpoints. Thus density also survives.

The missing point is a cut, not a pair of adjacent elements; the axioms do not require all cuts to be filled.