Does adding S->SS in a context-free grammar change the language to its Kleene star?
As chi pointed out in the comment, since $\varepsilon\in L^*$ and $\varepsilon$ may not belong to the new grammar, so adding $S\rightarrow SS$ does not always generate $L^*$. It makes more sense to ask whether it generates $L^+$, so the following answer focuses on $L^+$.



General Form

Consider the following grammar:

\begin{align}
S&amp;\rightarrow aSb\\
S&amp;\rightarrow \varepsilon
\end{align}

It generates the language $L=\{a^nb^n\mid n\in\mathbb{N}\}$.

Now after adding $S\rightarrow SS$ in this grammar, the string $aababb$ can be generated by the grammar but it does not belong to $L^+$. 

Chomsky Normal Form

Easy to see $L^+$ can be generated by the new grammar. Now for any string $s$ generated by the new grammar, we can prove $s\in L^+$ by mathematical induction on the depth of its parse tree. Consider the derivation process of $s$.

If the first derivation does not use the rule $S\rightarrow SS$, we can conclude $S$ does not show up in the following derivation since $S$ does not show up in the right hand side of a CFG of Chomsky normal form. As a result, $s\in L$. 

If the first derivation uses the rule $S\rightarrow SS$, the strings generated respectively by the two $S$'s in the right hand side must belong to $L^+$ due to inductive assumption, thus $s$, which is formed by concatenating the two strings, also belongs to $L^+$.