Are NP-complete sets formed from two other sets only if at least one is NP-hard? This question is somewhat of a converse to a previous question on sets formed from set operations on NP-complete sets:

If the set resulting from the union, intersection, or Cartesian
product of two decidable sets $L_1$  and $L_2$ is NP-complete, is at
least one of $L_1, L_2$ necessarily NP-hard?  I know that they cannot
both be in P (assuming P != NP) since P is closed under these set operations.  I also
know that the conditions of "decidable" and "NP-hard" are necessary
since if we consider any NP-complete set $L$ and another set $B$ outside
of NP (whether just NP-hard or undecidable) then we can form two
new NP-hard sets not in NP whose intersection is NP-complete.  For
example: $L_1:= 01L \cup 11B$, and $L_2:= 01L \cup 00B$.  However,
I don't know how to proceed after that.  

I'm thinking that the case of union might not be true since we can take a NP-complete set $A$ and perform the construction in Ladner's Theorem to get a set $B \in$ NPI which is a subset of $A$.  Then $B \cup (A \setminus B) =  A$ is the original NP-complete set.  However, I don't know if $A \setminus B$ is still in NPI or NP-hard.  I don't even know where to start for the case of intersection and Cartesian product.
The intersection of two non-NP-hard languages can be NP-hard.  Example: The solutions of any 3SAT instance are the set intersection of the solutions of a HORN-3SAT instance and an ANTIHORN-3SAT instance.  This is because a 3CNF clause must be either a Horn or anti-Horn clause and a 3SAT instance is the conjunction of such clauses.  3SAT is of course NP-complete; HORN-3SAT and ANTIHORN-3SAT are both in P.