Here are the rest of the 25 letter sets, found using brute force with the same 1 vowel per word with no repeated letters trick.

  [['brung', 'waqfs', 'vozhd', 'cylix', 'kempt'],
   ['brung', 'waqfs', 'vozhd', 'xylic', 'kempt'],
   ['jumby', 'waqfs', 'vozhd', 'clipt', 'kreng'],
   ['jumby', 'waqfs', 'vozhd', 'pling', 'treck'],
   ['jumby', 'waqfs', 'vozhd', 'prick', 'glent'],
   ['jumpy', 'waqfs', 'vozhd', 'bling', 'treck'],
   ['jumpy', 'waqfs', 'vozhd', 'brick', 'glent']]

I was a little inspired by this thread [0] about the practicality of implementing brute force algs in pure python for solving wordle. Using only built-ins and with a bit of optimisation [1] it runs in 4 minutes single threaded in cpython.

[0] https://twitter.com/eevee/status/1484716294179934209

[1] Instead of considering every possible combination, it discards non-optimal subsets. E.g. when considering o-words, only u- and a-word combos that already cover 10 letters are looked at. This significantly dampens the combinatorial explosion