$\newcommand\ep\varepsilon$Your stochastic domination condition $X\le_1-X+c$ can be written as 
$$P(X>x)\le P(-X+c>x)\tag{1}\label{1}$$
for all real $x$. 

If $c\ge2M$, then $X\le M\le2M-X\le-X+c$, so that $X\le-X+c$ and hence $X\le_1-X+c$. 

On the other hand, if $c<2M$ and $P(X=M)=1$, then 
$$P(X>c/2)=1\not\le0=P(X<c/2)=P(-X+c>c/2)$$
so that \eqref{1} does not hold for $x=c/2$ and hence it is not true that $X\le_1-X+c$. 

(If you insist that the support of $X$ be the entire interval $[-M,M]$, just use an approximation. For instance, you can assume that $P(X\in B)=(1-\ep)\,1(M\in B)+\ep\dfrac{|B\cap[-M,M]|}{2M}$ for any $\ep\in(0,1/2]$ and all Borel subsets of $\Bbb R$. Then the support of the distribution of $X$ will be the entire interval $[-M,M]$, whereas, for $c<2M$,  
$$P(X>c/2)>P(X=M)=1-\ep\ge\frac12$$
and hence
$$P(-X+c>c/2)=P(X<c/2)=1-P(X>c/2)<\frac12,$$
so that 
$P(X>c/2)>P(-X+c>c/2)$ and hence \eqref{1} does not hold for $x=c/2$. So, it is not true that $X\le_1-X+c$ if $c<2M$.)

So, the best lower bound on $c$ under the given conditions is $2M$. (There is no finite upper bound on $c$.