Arguing that $L= \{a^n | n\geq 0\} \cup \{a^nb^n| n\geq 0\}$ is not $LL(k)$ for any $k$
Rosenkrantz and Stearns prove in their paper Properties of deterministic top-down grammars that the language
$$ \{ a^n b^n : n \ge 0 \} \cup \{ a^n c^n : n \ge 0 \} $$
is not $\mathrm{LL}(k)$ for any $k$ (see page 246). Presumably a similar proof will show that your language is not $\mathrm{LL}(k)$ for any $k$.

You mention a confusion between grammars and languages. By definition, a language is $\mathrm{LL}(k)$ if it has an $\mathrm{LL}(k)$ grammar. To show that a language is not $\mathrm{LL}(k)$, you have to show that it has no $\mathrm{LL}(k)$ grammar.