Over a field, the answer to the first question is no.  An element of $A$ of order 4 in SL(2,R) must satisfy the equation $x^{4}-1$.  But it also satisfies the characteristic polynomial $x^{2}-{\rm Tr}(A) x+{\rm det}(A)$.  Taking the gcd of these two polynomials, we quickly see that $A$ also satisfies ${\rm Tr}(A)^{3}x+(1+{\rm Tr}(A)^{2}{\rm det}(A)-{\rm det}(A)^{2})$.  If ${\rm Tr}(A)\neq 0$ then $A$ satisfies a linear polynomial, so is diagonal, thus has 4th roots of unity along the diagonal.  There are only finitely many such matrices.  If ${\rm Tr}(A)=0$ then ${\rm det}(A)^{2}=1$, which says that $A$ is a root of $x^{2}+1$, which means any two elements of order 4 have equal squares (except for the finitely many possibilities expressed in the previous case).

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As for the original question: $S_{4}$ has a presentation $\langle s,t| s^{2}=t^{3}=(st)^{4}=1\rangle$.  I would suggest looking at the ring
$R=\mathbb{Z}[a,b,c,d,e,f,g,h]/I$ where $I$ is the set of relations forcing the 2x2 matrices give by s=((a,b),(c,d)) and t=((e,f),(g,h)) to satisfy the relations for $S_4$.  Finding a Grobner-type basis for $I$ should demonstrate that the resulting structure has at least 24 distinct matrices in the group generated by s and t (or it doesn't and you cannot embed $S_{4}$ in such a ring).