There are probably few or no "interesting" and "natural" closure properties for the class of languages that are not context-free. In fact that's probably true of:
any class of languages that is defined by a specific automaton, grammar or computational model -- the regular languages, the cfl's, various subclasses of the cfls such as the linear languages and the deterministic cfl's, the context-sensitive languages, bounded languages, and so on.
classes actually defined by closure properties, like the least family containing, say, {$a^n b^n$} and closed under the regular operations and transduction. In fact, AFL theory was all about things like that.
The reason is that many if not most or all "interesting" closure properties have the ability to drastically simplify a language, for example map it down to finite sets or something equally simple. For example, you can always apply a constant homomorphism (h(a) = 0) to a non-context-free language and get the language of all strings of zeros, which is context-free (in fact regular). So if the very definition of the class entails that it is not very "simple" (like non-context-free), then closure takes you down to "simple" languages, that is, outside the class.
This might actually make an interesting research project, part of which, I would hazard a guess, would be to define "simple", "interesting" and "natural" in some suitable way, and also to find formal ways to deal with trivial and degenerate cases like the one I gave.
The union and concatenation operations would be one test of such a theory. I'll conjecture that there are two non-cfl's which concatenate to $ \Sigma^*$ and others which union to that. I'd start with non-cfl's over a single letter, so you are really talking number theory and linear algebra.
Adendum -- here are two languages that definitely concatenate to $a^*$ and are almost certainly not context-free:
$\{a^{n-i}:i = \lfloor \sqrt n \rfloor\}$
$\{a^{n+i}:i = \lfloor \sqrt n \rfloor\}$
If for some reason, those turn out CF, use an even stranger function than $ \sqrt x $