Are there known "completion" operations over languages ?
Some observations:

1) For any NPI set $L$, there is an isomorphic set $L'$ and a set $S$ such that $L' \cup S$ is NP-complete.  Let $L' = \{0x : x \in L\}$ and let $S = \{ 1x : x \in SAT \}$.  This observation applies in a more direct manner to some of the natural suspected NPI sets (such as $GA$ -- see my comments on turkistany's answer).  It will also apply to the padded version of any set in $NEXP \backslash EXP$ (assuming $NEXP \neq EXP$, such a set is in NPI), for most reasonable ways of padding sets.

This rasises a potentially interesting question: 


  Is it the case that for any $L \in NPI$, there is an isomorphic language $L'$ and a set $S$ that is not NP-complete such that $L' \cup S$ is NP-complete?


2) R. Downey and L. Fortnow. Uniformly hard languages. Theoretical Computer Science, 298(2):303-315, 2003 (available from Lance's web page).  They show that the current proofs of Ladner's Theorem cannot easily be modified to produce uniformly hard languages (their definition of uniform hardness is natural, but is also in a sense aimed at Ladner-type constructions; it is not simply almost-everywhere hardness).  In particular, they show that if $P = PSPACE$ there is a minimal uniformly hard language under honest reductions; in particular, there is no relativizing proof of Ladner's Theorem for honest p-degrees that produces uniformly hard languages.  

In particular, that means that any relativizing proof of Ladner's Theorem that works for honest p-degrees (all known proofs fall in this category) produces sets with arbitrary polynomially-long stretches where they look like a set in $P$.  Now, that's not quite enough to be able to say that if $L$ is a language produced by such a construction then there is a set $S$ such that $L \cup S$ is NP-complete, but it's pretty close.