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Let $m=|A|$. By Singer's theorem, for every prime power $q$ there is a set $T\subset[1,L]\cap\mathbb Z$, where $L=q^2+q+1$ and $|T|=q+1$, with unique nonzero differences modulo $L$. Its ordinary nonzero differences are therefore unique. This implies strong Sidonicity: a nontrivial pair-sum equality would give two distinct ordered pairs with the same nonzero difference, also when a summand is repeated.

For each positive difference represented in $A$, delete one endpoint of the pair of $T$ representing that difference, if it exists. The remaining set $T'$ is strong Sidon, shares no positive difference with $A$, and has size at least $q+1-\binom m2$.

Take $q$ large enough that $L>N$, and put

$$B=(L+N)+T',\qquad M=2L+N.$$

Old-old sums are at most $2N$. Old-new sums lie between $L+N+2$ and $2L+2N$. New-new sums are at least $2L+2N+2$. These intervals are disjoint. Each of the old-old and new-new types is separately Sidon. An equality between two different old-new pairs would yield a common nonzero difference of $A$ and $T'$, which the deletions excluded. Thus $A\cup B$ is strong Sidon and

$$|A\cup B|\ge q+1+m-\binom m2 =\left(\frac1{\sqrt2}+o(1)\right)\sqrt M.$$

Here $A,N$ are fixed while $q$ tends to infinity through prime powers. This gives arbitrarily large $M$ for every fixed positive $\varepsilon$.