But in space the resistor doesn't have anywhere to put off the heat!
Actually, it does. Heat transfer can occur by three means: conduction, convection, and radiation. Very basically, heat conduction is about solid materials touching each other; convection is about gases or liquids touching the heat source; and radiation is about transmission of energy by means of releasing waves or particles. (This doesn't capture all of it, but since you are asking this question, I get the feeling that you aren't very familiar with the subject, and this is hopefully good enough to get you started for the purposes of this answer.)
In an atmosphere, convection is commonly a major mode of heat transfer. It's how every air cooled gadget (whether forced air cooling or ambient air) remains at an appropriate temperature, and it's mostly the way everything eventually ends up at the ambient temperature.
In space, there is no atmosphere, so convection doesn't work for cooling. But there's still conduction and radiation.
Conduction basically just means that if you leave your spacecraft somewhere far away from any heat source, or in an area of uniform heat sources surrounding it, everything within it will eventually have the same temperature. That's not particularly useful for our purposes; in a spacecraft, it's more about heat transfer within the spacecraft structure than to outside of it.
But even with convection and conduction not providing any useful heat transfer to keep our resistor cool, there is still radiation!
And in fact, that's how spacecraft maintain an appropriate temperature: By carefully controlling the heat and energy budget, not uncommonly ensuring that all sides of the spacecraft are exposed roughly equally over time to the heat source (which in our real world cases means the Sun) and matching heat dissipation against heat generation through radiation of excess heat.
For this reason, spacecraft designs include radiators which take heat generated and radiates it into space.
In that case, will the resistor keep heating?
Yes, unless the spacecraft includes radiators or some other way to dump excess heat; which it will, at least if it is intended to work for any length of time.
Will that change its resistance in turn affect the $U=IR$ relation?
Yes and no! This has been pointed out several times in comments, but I see no answer capturing it. Regardless of how exactly it is phrased, Ohm's law is valid only for a snapshot in time. This means that for $U=IR$ to hold as stated, you must simultaneously measure two or three of the quantities involved (voltage, current and resistance); if you measure two, you can calculate the third.
The voltage that is lost through resistance across the resistor becomes heat, which (unless it is somehow released) increases the temperature of the resistor.
Real-world resistors have a tendency to change their resistance when their temperature changes, which means that $R$ changes. In turn, either the voltage across the resistor ($U$), or the current through the resistor ($I$), must change for the equality $U=IR$ to remain valid. But if you were to measure these quantities again a microsecond later, you would find that the equality still holds, albeit with slightly different values for each.