Difference between $ L_1 = \{(a^n b^n)^m \mid n, m \ge 1\} $ and $ L_2 = \{a^n b^n \mid n \ge 1\}^+ $
Indeed, $abaaabbb \notin L_1$ because the string is not of the form $(a^nb^n)^m$ which is the repetition of a fixed string with the same number of $a$ and $b$.
The language $L_2$ is the Kleene closure of $\{a^nb^n \mid n\ge1\}$, consisting of all arbitrary concatenations of strings of the form $a^nb^n$. We can choose different strings of this form, and do not have to stick with the same one each time, like in $L_1$. Hence $L_2$ contains $abaaabbb$.
As for your original version of the question (now edited) it contained the notation $\{(a^nb^n)^+ \mid n\ge1\}$. I would not denote a language that way. The Kleene plus is a language operator, which takes a language (set of strings), and turns it into a language. By writing $\{(a^nb^n)^+ \mid n\ge1\}$ you get the set $(a^nb^n)^+$ within the set brackets, which can be called a &quot;type error&quot;.