Find all straight lines lying in S.

Either specification that I gave originally does in fact include lines other than those that go through the origin.  (Only two lines fitting the criteria actually pass through the origin: one coincident with the x-axis and one coincident with the y-axis.)

I had specified the intersection of the planes y=k;z=kx for any real k and x=k;z=ky for all real k.  That y=k, for values of k that differ from zero, precludes going through the origin, as y is zero at the origin.  Likewise for the case of x equals a non-zero k.

In my parametric form:
Set one: y=k;x=t;z=kt
Set two: y=t;x=k;z=kt
my k is your x0 in one instance and your y0 in the other.

It is true that all the lines I give pass through the x-axis or y-axis, and therefore look like they pass through the "origin" of a given x-z plane or y-z plane taken as a 2-dimensional cartesian system.  But this in fact represnts the actual case: no other straight lines exist on the surface.

In your solution set:
x=x0, y=y0+b*t, z=z0+c*t; where c=b*x0
x=x0+a*t, y=y0, z=z0+c*t; where c=a*y0
the implication is that y0 and z0 are independently assignable, but this is not the case while still fitting the original surface. On the surface, z0 must equal x0 times y0. Together with your caveat that c=b*x0, your first set becomes

x=x0, y=y0+b*t, z=x0*y0 + b*x0*t

So any given one of these lines still has x as a constant--x0 rather than k-- and the slope of z relative to y is b*x0/b, which is still x0, or my k, and when y is zero (so that y0=-b*t), z = x0*(-b*t) + b*x0*t, which is indeed equal to zero also.

So your set is the same as mine, once given the proper caveat that z0=x0*y0, which you failed to include.