He assures me that this is a documented feature.

Indeed it is, right at the top of perlref:

Note that taking a reference to an enumerated list is not the same as using square brackets--instead it's the same as creating a list of references!

    @list = (\$a, \@b, \%c);
    @list = \($a, @b, %c);      # same thing!
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As a special case, \(@foo) returns a list of references to the contents of @foo, not a reference to @foo itself. Likewise for %foo.

Update
I realized I didn't answer the question....

I think:
B::Deparse to the rescue: your code compiles to:

SWITCH: {
    last SWITCH if $id == $CASH and $tcash += $amt;
    last SWITCH if $id == $ACCOUNT and $taccount += $amt;
    last SWITCH if $id == $CHECK and $tcheck += $amt;
    last SWITCH if $id == $GIFT and $tgift += $amt;
    last SWITCH if $id == $VOUCHER and $tvoucher += $amt;
    last SWITCH if $id == $CC_MAN_AUTH and $tcc_man_auth += $amt;
    $tcredit += $amt;
}
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So you're continuing to the next case if $amt and the value being increased are both zero. Though that shouldn't cause any harm....

Update 2
Ack, I think I got it! Thanks Aristotle!
$amt can be negative. It can still bring one of the others to zero, then get decremented from the following branch.