Does a linear transformation of sets preserve subset properties? Suppose there are two sets $S_1$ and $S_2$ (not necessarily convex). Let $A$ be an arbitrary matrix, and say $AS_1$ is the set of vectors $y$ satisfying: 
$
y \in AS_1 \text{ if } y = Ax \text{ for some } x \in S_1. 
$
A similar definition holds for $S_2$.
If $AS_1 \subset AS_2$ , does it mean $S_1 \subset S_2$ for any matrix $A$ and any sets $S_1$ and $S_2$? or are there special cases?
 A: The answer is not specific to matrix multiplication.  If we have any function $f$ (from any set to any set) and subsets $S_1$ and $S_2$ of the domain, if $S_1 \subset S_2$ then $f[S_1] \subset f[S_2]$.  That is, pointwise application preserves the subset relation going forward. (Here $f[S]$ denotes the set $\{f(x) : x \in S\}$, which is the pointwise image of the set $S$ under the function $f$.)
What you are asking is the converse.  The answer is negative: if $f[S_1] \subset f[S_2]$ then it does not follow that $S_1 \subset S_2$.  It follows if $f$ is injective (one-to-one,) but if $f$ is not injective then you can always find a counterexample.  To see this, let $x_1$ and $x_2$ be distinct points with $f(x_1) = f(x_2)$.  Let $S_1 = \{x_1\}$ and $S_2 = \{x_2\}$.  Then the pointwise images are equal: $f[S_1] = \{f(x_1)\} = \{f(x_2)\} = f[S_2]$ but the sets $S_1$ and $S_2$ are distinct (and in fact neither one is a subset of the other.)
In your situation the function $f$ is multiplication by a matrix $A$; that is, $f(x) = A\textbf{x}$.  The same argument still applies.  Note that multiplication by $A$ represents an injective function if and only if the nullspace of $A$ is the trivial subspace $\{\textbf{0}\}$.  If the nullspace of $A$ is nontrivial, then one counterexample to your question (besides the general form of counterexample given above) is given by letting $S_1$ be the nullspace of $A$ and letting $S_2 = \{\textbf{0}\}$.
