First, you can always randomly sort the list in the end, so let's not worry about making "random permutations" (hard); and just worry about 1) making permutations (easy) and 2) randomizing them (easy).
If you want "truly" random groups, you have to accept that randomization by nature doesn't really allow for the constraint of "even distribution" of results -- you may get that or you may get a run of similar-looking ones. If you really want even distribution, first make the sets evenly distributed, and then randomize them as a group.
Do you have to use each element in the set x evenly? It's not clear from the rules that I couldn't just make the following interpretation:
Note the following: "over all the lists, the number of times each elements is used is the same (or as close as possible)"
Based on this criteria, and the rule that z < x*, I postulate that you can simply enumerate all the items over all the lists. So you automatically make y list of the items enumerated to position z. Your example doesn't fulfill the rule above as closely as my version will. Using your example of x={0,1,2,3} y=6 and z=2, I get:
0,1 0,1 0,1 0,1 0,1 0,1
Now I didn't use 2 or 3, but you didn't say I had to use them all. If I had to use them all and I don't care to be able to prove that I am "as close as possible" to even usage, I would just enumerate across all the items through the lists, like this:
0,1 2,3 0,1 2,3 0,1 2,3
Finally, suppose I really do have to use all the elements. To calculate how many times each element can repeat, I just take (y*z)/(count of x). That way, I don't have to sit and worry about how to divide up the items in the list. If there is a remainder, or the result is less than 1, then I know that I will not get an exact number of repeats, so in those cases, it doesn't much matter to try to waste computational energy to make it perfect. I contend that the fastest result is still to just enumerate as above, and use the calculation here to show why either a perfect result was or wasn't achieved. A fancy algorithm to extract from this calculation how many positions will be duplicates could be achieved, but "it's too long to fit here in the margin".
*Each list has the same z number of elements, so it will be impossible to make lists where z is greater than x and still fulfill the rule that no list may contain the same element twice. Therefore, this rule demands that z cannot be greater than x.