It is known that $SL_2(\mathbb{Z})= <S,R> $ where the generators are $S=\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}$ and $R=\begin{pmatrix} 1 & -1\\ 1 & 0 \end{pmatrix}$ of orders 4 and 6 resp.
Additionally it is known that- $$ PSL_2(\mathbb{Z})\cong <S | S^2>*<R|R^3>\cong <S,R | S^2, R^3> $$ the free product of cyclic groups of order 2 and 3. With the generators being the images of $S$ and $R$ in $PSL_2(\mathbb{Z})$
I hope to find a proof (possibly using the previous known facts) that- $$SL_2(\mathbb{Z}) \cong <S,R | S^4,R^6, S^2=R^3> $$
That is, isomorphic to an amalgamated product.
Hopefully it becomes easier when using the known facts...?
Let $G = \langle R,S \mid S^4=R^6=1, S^2=R^3 \rangle$,and let $N$ be its subgroup $\langle S^2 \rangle$. Then $N \le Z(G)$, $|N| \le 2$, and $G/N \cong \langle S,R \mid S^2=R^3=1 \rangle$.
Since the two matrices that you denoted by $S$ and $R$ satisfy the group relations and generate ${\rm SL}(2,\mathbb{Z})$, there is a surjective homomorphism $\tau:G \to {\rm SL}(2,\mathbb{Z})$.
We know that $\tau$ induces an isomorphism $G/N \to {\rm PSL}(2,\mathbb{Z})$. So, if $g \in \ker \tau$, then $g \in N$. But $\tau(N) = \left\langle \left( \begin{array}{cc}-1&0\\0&-1\end{array}\right)\right\rangle$ has order $2$ and $|N| \le 2$, so $\tau$ restricted to $N$ is injective. Hence $\tau$ is injective, and so it is an isomorphism, and $|N|=2$.