bash, 78 bytes (using a BSD utility) or 79 bytes (non-BSD also)
This is a little longer than @DigitalTrauma and @hvd's nice 71-byte bash solution, but I kind of liked the idea here of using numbers in base 60; I'm curious if anybody can golf this down a little more.
With the BSD-standard jot utility:
jot '-wx=`dc<<<60do3^%d+n`;((`dc<<<$x++'$1'-n`))||tr \ :<<<${x:3}' 86400 0|sh
With the more universally available seq utility:
seq '-fx=`dc<<<60do3^%.f+n`;((`dc<<<$x++'$1'-n`))||tr \ :<<<${x:3}' 0 86399|sh
The idea is to generate the numbers from 0 to 83699 and use dc to convert them into base 60. The "digits" in dc's base-60 output are 2-digit numbers from 00 to 59, with spaces separating the "digits", so this lists all the desired times from 00 00 00 to 23 59 59 in almost the format needed.
If you literally carry that out, though, numbers below 60^2 aren't 3-digit numbers in base 60, so the initial 00 or 00 00 is missing. For that reason, I'm actually generating the numbers from 60^3 to 60^3+83699; this ensures that all the numbers generated are exactly 4 digits long in base 60. This is OK as long as I eventually throw away the extra first digit (01) that isn't needed.
So, once the desired times are generated, I just take each quadruple from 01 00 00 00 to 01 23 59 59, add the last three numbers and subtract the argument $1. If that's 0, I then take everything in the quadruple from the 3rd character on (throwing away the "01 "), use tr to convert spaces to colons, and print the result.