Regular language with pumping lemma
Your concatenation idea doesn't work. Although the concatenation of two regular languages is guaranteed to be regular, the concatenation of a regular language and a non-regular language is not guaranteed to be non-regular. For example, take $L_1=\Sigma^*$, $L_2=\{a^nb^n\mid n\geq 0\}$. $L_2$ is not regular but $L_1L_2=\Sigma^*$ is regular.

To prove that $S$ is non-regular using the pumping lemma, pump a string that contains more $b$s than $c$s.