Just wanted to follow up with a short update:
I dissolved 5 g of Na2-EDTA in 100ml distilled water and added enough NaHSO4 to lower pH below 1. A precipitate quickly formed which could be filtered and washed with distilled water. After letting the filter plus its contents dry the filter weighed 4 grams more than a fresh filter from the same box. Since I wanted to know whether I created Na-EDTA, EDTA or EDTA anhydride with this method, I did the math to find out how much NaOH is needed to make 1g of each compound soluble as Na2-EDTA in 40 ml water. According to my calculations it takes 0.127g NaOH for Na-EDTA, 0.274g NaOH for EDTA, and 0.312g NaOH for EDTA anhydride.
A quick test confirmed that 0.186g NaOH (it comes in pellets, hence the odd numbers) could not, but that 0.289g could slowly dissolve the 1g of unknown powder in 40ml water, which tells me that most likely EDTA free acid was formed in the reaction with NaHSO4. I also did some pH measurements which seem to confirm this:
1.35 g Na2-EDTA in 40 ml water: pH = 4.28
0.186g NaOH + 1g test powder in 40 ml water: pH = 3.60
0.289g NaOH + 1g test powder in 40 ml water: pH = 4.35
This tells me that with only slight excess NaOH (0.289g instead of the calculated 0.274g for EDTA) we get only a slight rise in pH.
With the evidence at hand I would conclude that I produced about 4 grams of EDTA free acid from 5 grams of Na2-EDTA and quite a bit of NaHSO4. Availability and cost will be determining factors whether this is a good starting point for making Ammonium Ferric EDTA, but from a technical standpoint it works.