If $a,b,c,d \in S$ and $a+b=c+d$ gives $a=c$ or $a=d$ then let us call $S$ a Sidon set. Suppose such $M$ exists. By the second condition, $S= M \setminus \{1, \cdots, 10\}$ is a Sidon set. Let $S(x)$ be the number of elements of $S$ less than or equal to $x$. For any $a,b,c,d \in S$ such that $a,b,c,d \le x$ and $a>b$ and $c>d$ and $(a,b)\neq (c,d)$, we have $a-b \neq c-d$ since $S$ is a Sidon set. Since the $\binom{S(x)}{2}$ such pairs all have distinct differences below $x$, we get $\binom{S(x)}{2} \le x$ and $S(x) < \sqrt{2x}+1$. On the other hand, $\mathbb{Z}_{>1} = M+M = (S+S) \cup (S+\{1,\cdots,10\}) \cup \{2, \cdots, 20\}$. Let us cound the number of elements in this set not exceeding $x$. There are exactly $x-1$ elements on the left side, and at most $\binom{S(x)}{2} + 10 S(x) + 19$ elements on the right. Hence $x \le \binom{S(x)}{2} + 10S(x) + 19$ and therefore $S(x) > \sqrt{2x} - O(1)$. (Actually there are more tighter bounds http://en.wikipedia.org/wiki/Sidon_set) Now $S(x) = \sqrt{2x}+O(1)$. Let $A = S \cap [1,x]$ and $B = S \cap [x+1,2x]$. Then $|A| = \sqrt{2x}+O(1)$ and $|B|=(2-\sqrt{2})\sqrt{x} + O(1)$. The number of positive elemtents of $A-A$ is $\binom{|A|}{2} = x + O(\sqrt{x})$ and the number of positive elements of $B-B$ is $\binom{|B|}{2} = (3-2\sqrt{2})x + O(\sqrt{x})$. All of these elements do not coincide since $S$ is a Sidon set and are below $x$. Hence there exists $(4-2 \sqrt{2})x + O(\sqrt{x})$ numbers below $x$, and this leads to a contradiction when $x$ is sufficiently large. Therefore such $M$ does not exist.