Proving a LL(1) equivalent grammar doesn't exist
The language for the grammar in the question is the set of all words with equal number of $a$'s and $b$'s. 

I am afraid your intuition is incorrect. Here is an $LL(1)$ grammar for it.

$S \to aAbS \mid bBaS \mid \epsilon$
$A \to aAb \mid \epsilon$
$B \to bBa \mid\epsilon$

Here is some statistics about the above $LL(1)$ grammar.
$$\begin{array} {|c|c|c|c|}
\hline
\text{nonterminal} &amp;\text{first set} &amp;\text{follow set} &amp;\text{nullable} &amp;\text{endable}\\\hline
S &amp;a\ b &amp;\emptyset &amp;\text{yes} &amp;\text{yes}\\\hline
A  &amp;a &amp;b &amp;\text{yes} &amp;\text{no}\\\hline
B &amp;b &amp;a &amp;\text{yes} &amp;\text{no}\\\hline
\end{array}$$

Here is the parsing table.
$$\begin{array} {|c|c|c|c|}\hline
 &amp;a &amp; b &amp;\$ \\\hline
S  &amp;S\to aAbS &amp;S\to bBaS &amp;S\to \epsilon\\\hline
A  &amp;A\to aAb &amp;A\to \epsilon\quad\ &amp; \\\hline
B &amp;B\to \epsilon\quad\ &amp;B\to bBa &amp; \\\hline
\end{array}$$





Exercise. Show that both the grammar in the question and the grammar in this answer generate the language of words with equal number of $a$'s and $b$'s.