Here is my proof for the uniqueness statement above. $\hat{f}$ denotes the fourier transform of $f\in L^2.$
(i) (the if part):
Let $g \in M_a.$ Then $m(G\cap A)=0,$ where $G=\{t\in \mathbb{R} | \hat{g}(t)\neq0 \}.$
We now have to show that $m(G\cap B)=0,$ so that $g\in M_B.$
But, $m(G\cap B)= m(G \cap(B-A))+m(G\cap(B\cap A))\leq m(B-A) +m(G\cap A) = 0.$
The other case $M_A \subset M_B$ also follows by symmetry.
(ii)(the only if part):
Suppose that $M_A=M_B.$ Then, $M_A \subset M_B$ and $M_B \subset M_A.$ - (1)
Note then that for any Lebesgue measurable set $F \subset \mathbb{R^1},$ we can take $f\in L^2$ so that $\hat{f}\neq0$ on $F,$ because the Fourier transform is a homeomorphism on $L^2$, and thus $e^{-|x|}\cdot1_F \in L^2$ can always be chosen to be such $\hat{f}.$
Therefore, (1) can be rephrased as $m(F\cap A)=0 $ implies $m(F \cap B) = 0$ & $m(F \cap B) = 0$ implies $m(F \cap A) =0.$
Now, let $F=\mathbb{R}.$ Then, we have that $m(A)=0 $ implies $m(B) = 0$ & $m(B) = 0$ implies $m(A) =0.$ -(2)
Thus, $m(A-B)>0$ contradicts the second one of (2) and $m(B-A)>0$ contradicts the first one of (2).
It therefore follows that $m((A-B)\cup (B-A)) = 0.$
I have corrected my proof much. Any comment rectifying possible errors or even another simpler or more correct proof would be appreciated.