I would have been surprised if your last example printed 'yes', but to be honest I wouldn't have known for sure without trying.
Technically, it's the same as number 7, besides the unnecessary setting of the (uninitialized) value of $_ to itself.
Your last example and number 7 do not try to set the key within the subroutine.
Both examples are working on the alias @_ provides and not directly on the hash key as in your next to last example.
Maybe you've been bitten by the similarity number 6 provides, where setting the key (through the alias $_) to the undef value $_ provides will vivify the key.
Whereas in your last example trying to reset the value of $_ to $_ will not vivify the key because you are not using the key by trying to initialize the key value with a defined value even if the value is undef.
Setting $_ to $_ ($_ = $_;) will not change the value, hence there will be no update which could vivify the key.
Changing the value of $_ (e.g. $_ = undef; or $_ = 1;) will cause an update which will vivify the key.
The interesting part in this issue is that $_ = $_; looks like it will be handled like a no-op without updating the internal stack or whatever you may call it what is going on behind the scene.