There is a cheap way via Cauchy's inequality. I will only do it for $a=1$ just to outline the idea. $$\frac{x}{2} (1+o(1) ) \leq \sum_{n \leq x } 1_A(x)= \sum_{n \leq x } 1_A(x)\sqrt{\tau(n)} \tau(n)^{-1/2}\leq (S_1^A(x))^{1/2} (\sum_{n\leq x } \tau(n)^{-1} )^{1/2} .$$ Now use $$ \sum_{n\leq x } \tau(n)^{-1} \ll \frac{x}{\sqrt{\log x } }$$ to get $$ M \gg x (\log x)^{1/2}.$$ This is a long way from $M\gg x \log x$ but it is better than the trivial bound $M \geq x/2 (1+o(1) ) $. You might try to use H"older's inequality instead to find a better logarithmic exponent.

There is a cheap way via Cauchy's inequality in case you do no want to use Erdos--Kac. I will only do it for $a=1$ just to outline the idea. $$\frac{x}{2} (1+o(1) ) \leq \sum_{n \leq x } 1_A(x)= \sum_{n \leq x } 1_A(x)\sqrt{\tau(n)} \tau(n)^{-1/2}\leq (S_1^A(x))^{1/2} (\sum_{n\leq x } \tau(n)^{-1} )^{1/2} .$$ Now use $$ \sum_{n\leq x } \tau(n)^{-1} \ll \frac{x}{\sqrt{\log x } }$$ to get $$ M \gg x (\log x)^{1/2}.$$ This is a long way from $M\gg x \log x$ but it is better than the trivial bound $M \geq x/2 (1+o(1) ) $. You might try to use H"older's inequality instead to find a better logarithmic exponent.