Certainly $\mathrm{SL}(2,\mathbb{Z})$ contains a free group. For instance $\Gamma(2)$, the subgroup of all matrices congruent to the identity modulo $2$, is free of rank $2$. The matrices $\left(\begin{array}{cc}1&2\\ 0&1\end{array}\right)$ and $\left(\begin{array}{cc}1&0\\ 2&1\end{array}\right)$ freely generate $\Gamma(2)$. This can be proved by considering the action on the upper half-plane or by careful examination of reduced words. There's a nice proof in chapter 18 of David Ullrich's book Complex Made Simple.
Your $a$ and $b$ don't generate a free group alas, since they generate all of $\mathrm{SL}(2,\mathbb{Z})$.
Re the edited question. Let's write $$T=\left(\begin{array}{cc}1&1\\ 0&1\end{array}\right)\qquad \textrm{and}\qquad U=\left(\begin{array}{cc}1&0\\ 1&1\end{array}\right).$$ As both Jack and I pointed out, $T^2$ and $U^2$ generate a free subgroup of rank $2$. Now it's an easy exercise to prove that $T$ and $U$ freely generate a free monoid of rank $2$ (because their entries are non-negative). On the other hand, they generate thw whole group $\mathrm{SL}(2,\mathbb{Z})$ which is certainly not free. Your matrices $a$ and $b$ are, if my calculations are right, $-U^{-1}$ and $-T^{-1}$. The matrix $S_4$ conjugates $T$ and $U$ into $U^{-1}$ and $T^{-1}$ so $U^{-1}$ and $T^{-1}$ freely generate a free monoid of rank $2$. The same must be true of $U^{-1}$ and $T^{-1}$, that is, of $a$ and $b$.
