Does the following transformation preserve context-freeness?
Counter-example:

$L_1 = \{a^n b^n c^m \mid m,n \ge 1 \}$

$L_2 = \{a^m b^n c^n \mid m,n \ge 1 \}$

$L = (L_1 \cdot L_2^*) \cup \epsilon$ is context-free.

Any nonempty word $x \in L^\#$ has a prefix $p = a^n b^n c^m \in L_1$. It must be $n = m$, because due to $L_2$, any pair of a $b^+$ and a directly succeeding $c^+$ in $x$ (after $p$) must share the same exponent. Therefore:

$L^\# = (\{a^n b^n c^n \mid n \ge 1 \} \cdot L_2^*) \cup \epsilon$, which is not context-free.