The identity is the only surjection of the positive integers onto themselves sending every sum-free set onto a sum-free set (no verification is needed, of course).
To prove this, fix a function $f$ satisfying the conditions in the statement, and proceed in several steps.

Step 1. Notice that a 2-element set $\{x, y\}$, where $x < y$, is not sum-free if and only if $y = 2x$.

Choose any $a \in \mathbb{N}$, and for any $i \ge 0$ choose some $x_i$ such that $f(x_i) = 2^i a$. The set $f(\{x_i, x_{i+1}\})$ is not sum-free, so neither is $\{x_i, x_{i+1}\}$, whence $x_i = 2x_{i+1}$ or $x_{i+1} = 2x_i$. Since the $x_i$ are all distinct, the same option should hold for all $i$. The former option yields $x_i = x_0 2^{-i}$ which cannot hold for large enough $i$. So $x_{i+1} = 2x_i$ for all $i$.
Therefore, $f(2x) = 2f(x)$ for all $x$, and, moreover, $x$ is the only argument $t$ with $f(t) = f(2x)/2$. Therefore, $f$ is injective (and hence bijective).

Step 2. Say that a 3-element set $\{a, b, c\}$ is good if it is not sum-free, but each of its 2-element subsets is (in other words, no element is twice another). It is easily seen that a set $\{a, b, c\}$, where $a < b$, is good only if $c = b \pm a$. Notice that the pre-image of a good set is also a good set, due to Step 1.
Now let $f(1) = a$. We show that $f(n) = an$ by induction on $n$. The base cases are $n=1, 2, 3, 4, 5$; for $n=1, 2, 4$ the result follows from Step 1.
Set $t = f^{-1}(3a)$ and $s = f^{-1}(5a)$. The sets $\{a, 4a, 3a\}$ and $\{a, 4a, 5a\}$ are good, hence so are $\{1, 4, t\}$ and $\{1, 4, s\}$. Therefore, $\{s, t\} = \{3, 5\}$. But the set $\{a, 5a, 6a\}$ is also good, so the pair $\{1, s\}$ is contained in one more good set, which is not the case if $s=3$, since $\{1, 3\}$ is contained in one single good set, namely, $\{1, 4, 3\}$. Thus $t=3$ and $s=5$, which establishes the base.
For the induction step, assume that $f(k) = ak$ for all $k \le n$, where $n \ge 5$. Choose $t = f^{-1}((n+1)a)$. Then the pair $\{a, na\}$ is contained in two good sets, namely, $\{a, na, (n-1)a\}$ and $\{a, na, (n+1)a\}$. Their pre-images, $\{1, n, n-1\}$ and $\{1, n, t\}$, are also good, and injectivity of $f$ forces $t = n+1$. This completes the induction step.
Finally, since $f$ is surjective, $1 = f(n) = an$ for some positive integer $n$, so $a=1=n$. Consequently, $f$ is the identity, as claimed at the beginning of the solution.

The approach is similar to Solution 1, but avoids directly defining and working with good sets. Step 1 in Solution 1 is proved similarly.
For any odd integer $a \ge 3$, we claim $f(a) + f(1)$ is equal to one of $f(a+1)$ and $f(a-1)$. Indeed, its preimage should, together with $a$ and $1$, form a set that is not sum-free. $f(a)+f(1) = f(2a) = 2f(a)$ contradicts injectivity, as does $f(a)+f(1) = f(2) = 2f(1)$, hence $f(a)+f(1) = f(a+1)$ or $f(a)+f(1) = f(a-1)$.

$f(2) = 2f(1)$, contradicting injectivity. Therefore, $f(a) + f(1) = f(a + 1)$ for all odd integers $a$.
Finally, we prove $f(n) = nf(1)$ by induction. Indeed, assume the statements holds for all integers up to some even $n$. The base case for $n \in \{1, 2\}$ holds by Step 1 of Solution 1. Then, by assumption,
$$f(n + 1) + f(1) = f(n + 2) = 2f\left(\frac{n+2}{2}\right) = 2\left(\frac{n+2}{2}\right)f(1) = (n+2)f(1),$$
implying $f(n + 1) = (n + 1)f(n)$ and, by the above result,
$$f(n + 2) = f(n + 1) + f(1) = (n + 2)f(1),$$
completing the induction.

Assume that $f(a)+f(1) = f(a-1)$ for some odd integer $2$. Inducting backwards, we see the statement holds for all odd integers up to $a$, in particular $f(3)+f(1) =$

The conclusion now follows as in Solution 1.