It is well know that the group $SL_2(\mathbb{Z})$ is generated by the matrices $S= \begin{bmatrix} 0 & -1 \\ 1 & 0\end{bmatrix}$ and $T= \begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}$. In particular, we have that
$SL_2(\mathbb{Z})= \langle S,T \mid S^4, ()^3=^2 \rangle $.
The presentation implies that for every $\gamma \in SL_2(\mathbb{Z})$ there exist a finite set $\{n_i,m_i\}_{I=1}^{k}$ with $n_i,m_i \in \mathbb{Z}$ such that
$\gamma = S^{n_1}T^{m_1} \cdots S^{n_k}T^{m_k}$.
Without loss of generality, we can assume that $n_i \in \{1,3\}$ for $i\neq 1$ and $0 \le n_1 \le 3$. Similarly, we can assume that $m_i \neq 0$ for $i\neq k$.
Under which assumptions, if there are any, we can assume the above factorization of $\gamma$ unique?
Just to be precise, let $\gamma \in SL_2(\mathbb{Z})$ be written in the two following way
$\gamma = S^{n_1}T^{m_1} \cdots S^{n_k}T^{m_k}, \quad \gamma=S^{\bar{n_1}}T^{\bar{m_1}} \cdots S^{\bar{n_k}}T^{\bar{m_k}}.$
We say that a factorization of the element $\gamma$ is unique we can conclude that $n_i = \bar{n}_i$ and $m_i = \bar{m}_i$.
I don't think there is unique factorization since there is one way to think of $S$ and $T$ generating $SL_2(\mathbb{Z}$ is by doing the Euclidean algorithm, and since we can either do the classic Euclidean algorithm or a signed Euclidean algorithm, this should lead to $2$ different factorizations. Since we have that $$ S\begin{bmatrix} a & b\\ c & d \end{bmatrix}=\begin{bmatrix}-c & -d\\a & b\end{bmatrix} $$ while $$ T^m\begin{bmatrix}a & b\\c &d \end{bmatrix}=\begin{bmatrix}a+mc & b+md\\ c&d\end{bmatrix} $$
For example, let us consider $$ X=\begin{bmatrix} 5 & 3\\ 3 & 2 \end{bmatrix} $$ Then we can observe that doing the classical Euclidean algorithm we have that $$ T^{-1}\begin{bmatrix} 5 & 3\\ 3 & 2 \end{bmatrix}=\begin{bmatrix}2 & 1\\ 3 & 2\end{bmatrix} $$ $$ ST^{-1}\begin{bmatrix} 5 & 3\\ 3 & 2 \end{bmatrix}=\begin{bmatrix}-3 & -2\\ 2 & 1\end{bmatrix} $$ $$ T^2ST^{-1}\begin{bmatrix} 5 & 3\\ 3 & 2 \end{bmatrix}=\begin{bmatrix}1 & 0\\ 2 & 1\end{bmatrix} $$ $$ ST^2ST^{-1}\begin{bmatrix} 5 & 3\\ 3 & 2 \end{bmatrix}=\begin{bmatrix}-2 & -1\\ 1 & 0\end{bmatrix} $$ $$ T^2ST^2ST^{-1}\begin{bmatrix} 5 & 3\\ 3 & 2 \end{bmatrix}=\begin{bmatrix}0 & -1\\ 1 & 0\end{bmatrix}=S $$ From this we get one factorization of $$ X=TS^3T^{-2}S^3T^{-2}S=TST^{-2}ST^{-2}S $$ Now we can give another factorization doing the Euclidean algorithm as $$ T^{-2}\begin{bmatrix} 5 & 3\\ 3 & 2 \end{bmatrix}=\begin{bmatrix}-1 & -1\\ 3 & 2\end{bmatrix} $$ $$ ST^{-2}\begin{bmatrix} 5 & 3\\ 3 & 2 \end{bmatrix}=\begin{bmatrix}-3&-2\\-1 & -1\end{bmatrix} $$ $$ T^{-3}ST^{-2}\begin{bmatrix} 5 & 3\\ 3 & 2 \end{bmatrix}=\begin{bmatrix}0&1\\-1 & -1\end{bmatrix} $$ $$ ST^{-3}ST^{-2}\begin{bmatrix} 5 & 3\\ 3 & 2 \end{bmatrix}=\begin{bmatrix}1&1\\0 & 1\end{bmatrix}=T $$ Thus, this calculation gives us $$ X=T^2S^3T^3S^3T=T^2ST^3ST $$ which is a different factorization.