Let's consider different interpretations of what "travel over" a place means to get a ballpark idea.
The ISS is literally "over" a given place:
The ISS is about 108.5 meters wide, at an altitude (recently) of about 400 km, and the Earth's equatorial radius (where the coverage is most sparse) is 6378 km. So on the surface that projects down to a 102 meter wide "footprint" on the surface at the equator. If each equator-crossing is at an angle of 51.6 degrees with respect to the equator, it paints a section of the equator that is 102 meters / cos(90-51.6) ~ 130 meters.
Since the Earth's equator is about 40 million meters in circumference, it would take an absolute minimum of 308,000 crossings to paint the equator. At a period of 93 minutes, that would be about 19,900 days or a minimum of 57 years.
That lower limit is unrealistic. Since the ISS isn't tightly coupled to the Earth's surface in a repeat ground-track orbit one could try a statistical analysis assuming that over the decades the passes were distributed randomly. In that case it turns out only about 63% of the equator's circumference would be painted in the first 57 years, and something like 63% of the leftover would be painted in the next 57 years, etc. (see python script at the end)
The ISS passes within 1 degree of the zenith:
That's about 7 kilometers, so the equator would be 63% painted after 5,700 orbits (about one year).
The ISS passes within 45 degrees of the zenith:
That's about 800 kilometers, so the equator would be 63% painted after 50 orbits (three days).
Remember this is just statistical, so there could be much longer periods where some areas go without seeing the ISS, but only very careful (and pointless) planning could prevent any one given place on the equator from not having the ISS within 45 degrees of the zenith within a given year.
note: for points above the equator, but within 51.6 degrees North and South latitude, the track's motion is increasingly parallel to the ground, so the probabilities become higher. Equator coverage is the hardest.
Here's a quick python calculator to confirm the "about 63%" statistical estimate:
import numpy as np
total_meters = int(6378 * 1000 * 2 * np.pi)
n_orbits = int(total_meters / 130)
equator = np.zeros(total_meters, dtype=bool)
positions = (total_meters * np.random.random(n_orbits)).astype(int)
for pos in positions:
equator[pos:pos+130] = True # ignore wrapping - can use np.put()
print "FYI: 1-exp(-1) = ", 1 - np.exp(-1)
gives for example:
FYI: 1-exp(-1) = 0.632120558829