So I thought I'd answered my own question (How close is the ISS to a closed system?), but I just realized there's a missing piece.
This is the chemical equation for how hydrogen, oxygen, and carbon are recycled and/or consumed on the station (now that the Sabatier system is in place).
$$2H_{2}O \overset{hydrolysis}{\rightarrow} 2H_2 + O_2 + \overset{food}{C} \overset{respiration}{\rightarrow} CO_2 + 2H_2 + \overset{added}{2H_2} \overset{sabatier}{\rightarrow} 2H_{2}O + \overset{discarded}{CH_4}$$
Removing the recycled components, it simplifies to:
$$\overset{food}{C} + \overset{added}{2H_2} \rightarrow \overset{discarded}{CH_4}$$
So the oxygen is recycled almost completely, but it still requires input hydrogen. This is provided by shipments of water. I thought recycling the oxygen saves a lot of water weight they don't have to launch anymore. But then I realized that if the hydrogen still comes from water, every two atoms of hydrogen arrives with an atom of oxygen. That's a lot of dead weight just to carry hydrogen to the station: sending 1kg of hydrogen requires launching 9kg of water.
And since oxygen isn't needed or consumed, where does it go? Presumably it's scrubbed out of the atmosphere as $CO_2$ and vented overboard. But then we're back where we started before Sabatier.
Basically, as far as I can figure it, the simplified equation went from this, before Sabatier:
$$2H_{2}O \overset{hydrolysis}{\rightarrow} 2H_2 + O_2 + \overset{food}{C} \rightarrow \overset{discarded}{CO_2 + 2H_2}$$
To this, after:
$$2H_{2}O \overset{hydrolysis}{\rightarrow} 2H_2 + O_2 + \overset{food}{C} \rightarrow \overset{discarded}{CH_4} + \overset{???}{O_2}$$
How is the water input reduced?