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 $x^{2}-{\rm Tr}(A) x+1$.  Taking the gcd of these two polynomials, we quickly see that $A$ also satisfies $({\rm Tr}(A)^{3}-2{\rm Tr}(A)x+{\rm Tr}(A)^{2}$.  If the coefficient of $x$ is nonzero 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 the coefficient of that linear polynomial is zero then ${\rm Tr}(A)=0$ and so $A$ 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 $S_4$ and have determinant 1.  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).