[basic.link]/8 in N4140 contains the following statement:
A type without linkage shall not be used as the type of a variable or function with external linkage unless
(8.7) — the entity has C language linkage (7.5), or
(8.8) — the entity is declared within an unnamed namespace (7.3.1), or
(8.9) — the entity is not odr-used (3.2) or is defined in the same translation unit.
Clearly a condition satisfying (8.8) is not possible, as an entity declared in an unnamed namespace cannot have external linkage, at the same time.
Then I decided to find an example of a function with external linkage returning a type with no linkage, i.e., returning an object of a local class, to no avail, irrespective whether the function was, or wasn't, in the same TU as the type. I'm afraid the set of options satisfying (8.9) may also be empty. If that's the case, I'd like to hear some confirmation on this.
In reference to (8.7) I don't know what to say, but it seems to me that this bullet point will not add anything new to the problem either.
One case is if you have templates.
template<typename T>
void f(T t) { t(); }
int main() { f([]{}); }
The instantiated function has external linkage, while T is a type without linkage. Strictly speaking, though, the instantiation is not instantiated into a TU (instantiations live in instantiation units, really). But I suspect that text is meant to apply here aswell.
As does this (I assume you already know, but this of course is a variable)
struct { } x;
With that in place, you can construct normal functions that satisfy the rule aswell
decltype(x) f() { return {}; }