The triangle $ABC$ is uniquely determined by the acute triangle $XYZ$ inscribed in $(S)$, where $X,Y,Z$ are the points where $BC,CA,AB$ touch $(S)$. Now let $M,N,P$ be the midpoints of $YZ,ZX,XY$ respectively. We have $SM\cdot SA=SN\cdot SB=SP\cdot SC=1$, so the ordering of $SA,SB,SC$ implies $SM\ge SN\ge SP$. If we find the loci of $M,N,P$, then the loci of $A,B,C$ will be the images of those (respectively) through the inversion wrt $(S)$. The condition $SM\ge SN\ge SP$ says that, in the acute triangle $XYZ$, we have $\angle X\le \angle Y\le \angle Z$. We can find acute triangles of the form $\angle X<\angle Y=\angle Z<\frac \pi 2$ with $\angle X$ taking any value in the interval $(0,\frac \pi 3]$, so $M$ is a the midpoint of a chord subtending an angle of no more that $\frac \pi 3$, which means that the locus of $M$ is the region $\frac 12\le SM<1$, so the locus of $A$ is the region $1<SA\le 2$. Regarding $\angle Z$, this is the largest angle, so it must be in the interval $[\frac \pi 3,\frac \pi 2)$. We can easily find triangles with their largest angle taking any of these values, so the locus of $P$ is the region $0<SP\le \frac 12$, meaning that the locus of $C$ is $2\le SC$. $\angle Y$ is the middle angle, and we have to find bounds for it. It can take values as close to $\frac \pi 2$ as we want (we take isosceles triangles s.t. $\angle Y=$ one of the other angles), but it must be $>\frac \pi 4$, because otherwise the largest angle would be $\ge \frac \pi 2$. The same reasoning shows that it can take values as close to $\frac \pi 4$ as we want, so the locus of $N$ is $0<SN<\frac{\sqrt 2}2$, so the locus of $B$ is $\sqrt 2<SB$.