That's quite the intention actually. The motivation here is that there is clearly a 1-to-1 and onto (bijective) mapping between these sets. While we can set a clear map even from Z to 2Z (T: x -> 2x) there are other maps that work perfectly as well. For example if we take the ideal/subset of Z such that we use every other number (i.e. the evens) then this also corresponds to 2Z but we still have an infinite number of numbers left over.

Let's be more clear

Z = {Z+} + {Z-} + {0} = {2Z+} + {(2n+1)Z+} + {2Z-} + {(2n+1)Z-} + {0}

2Z = {2Z+} + {2Z-} + {0}

Then if we do Z - 2Z we get {(2n+1)Z+} + {(2n+1)Z-} (or the odds without 0). In other words, we have infinity plus infinity but one of the infinities is fully contained in the other infinity. It does seem reasonable to think that these may not be the same size, just as we would say that the uncountable infinities is larger than the countable infinities. In fact, the logic is quite similar here.

Note though, that the sibling comment (to you) gave a counter example as to where there might still be issues. Because by this setup we'd still have the number of primes equal to the size of even integers. So I asked about rates of going to infinity.