1. Initial Observation: We need to determine all surjective functions $f: \mathbb{N} \to \mathbb{N}$ such that for any sum-free subset $A \subseteq \mathbb{N}$, the image $\{f(a) \mid a \in A\}$ is also sum-free.

2. Identity Function: The function $f(x) = x$ is a candidate. If $A$ is sum-free, then $\{f(a) \mid a \in A\} = A$ is also sum-free. We need to show that this is the only function that satisfies the condition.

3. Claim 1: For all $x \in \mathbb{N}$, $f(2x) = 2f(x)$. If $f(y) = 2f(x)$, then $y = 2x$.

Proof: Suppose $f(x) = c$. Since $f$ is surjective, there exists $y$ such that $f(y) = 2f(x) = 2c$. Thus, $f(x) + f(x) = f(y)$. If $\{x, y\}$ is sum-free, then $\{f(a) \mid a \in \{x, y\}\}$ must be sum-free, but $f(x) + f(x) = f(y)$ contradicts this. Therefore, $y = 2x$ or $2x = y$. Assume $y = \frac{x}{2}$. Then $f\left(\frac{x}{2}\right) = 2c$. By surjectivity, there exists $y_1$ such that $f(y_1) = 4c = 2f\left(\frac{x}{2}\right)$. By the same logic, $y_1 \in \left\{2 \cdot \frac{x}{2}, \frac{\frac{x}{2}}{2}\right\} = \left\{x, \frac{x}{4}\right\}$. Since $y_1 \neq x$, $y_1 = \frac{x}{4}$. Continuing this process, we get a sequence $y_k$ such that $f(y_k) = 2^{k+1}c$ and $y_k = \frac{x}{2^k}$. This is impossible since $\nu_2(x)$ is finite. Thus, $y = 2x$ and $f(2x) = 2f(x)$.

4. Claim 2: Let $\{c_i\}_{i \geq 1}$ be a sequence of naturals such that $f(c_i) = i$ for all $i \geq 1$. Then $c_i = ic_1$ for all $i \geq 1$.

Proof: The $c_i$ are pairwise distinct since if $c_i = c_j$, then $i = f(c_i) = f(c_j) = j$. We use induction on $i$. For $i = 1$, the result is trivially true. For $i = 2$, by Claim 1, $f(c_2) = 2 = 2f(c_1)$, so $c_2 = 2c_1$. For $i = 3$, $f(c_3) + f(c_1) = f(c_4)$, and from Claim 1, $f(c_4) = 2f(c_2)$, so $c_4 = 2c_2 = 4c_1$. Thus, $\{c_1, c_3, c_4\} = \{c_1, c_3, 4c_1\}$ is not sum-free. Since $\{c_1, 4c_1\}$ is sum-free, $c_3 \in \{\frac{c_1}{2}, 2c_1, \frac{4c_1}{2}, 2(4c_1), 4c_1 - c_1, 4c_1 + c_1\} = \{\frac{1}{2}c_1, 2c_1, 3c_1, 5c_1, 8c_1\}$. If $c_3 = \frac{1}{2}c_1$, then $f(c_1) = f(2c_3) = 2f(c_3) = 6$, absurd. If $c_3 = 2c_1$, then $f(c_3) = f(2c_1) = 2f(c_1) = 2$, absurd. If $c_3 = 8c_1$, then $f(c_3) = f(8c_1) = 2f(4c_1) = 4f(2c_1) = 8f(c_1) = 8$, absurd. Thus, $c_3 \in \{3c_1, 5c_1\}$. If $c_3 = 5c_1$, then $f(c_5) = f(c_1) + f(c_4) = f(c_1) + f(4c_1)$, so $\{c_1, 4c_1, c_5\}$ is not sum-free. Thus, $c_5 \in \{\frac{c_1}{2}, 2c_1, \frac{4c_1}{2}, 2(4c_1), 4c_1 - c_1, 4c_1 + c_1\} = \{\frac{1}{2}c_1, 2c_1, 3c_1, 5c_1, 8c_1\}$. If $c_5 = \frac{1}{2}c_1$, then $f(c_1) = f(2c_5) = 2f(c_5) = 10$, absurd. If $c_5 = 2c_1$, then $f(c_5) = f(2c_1) = 2f(c_1) = 2$, absurd. If $c_5 = 8c_1$, then $f(c_5) = f(8c_1) = 8f(c_1) = 8$, absurd. Thus, $c_5 \in \{3c_1, 5c_1\}$. $c_5$ cannot be $5c_1$ since we are assuming for the sake of contradiction that $c_3 = 5c_1$, and we know $c_3 \neq c_5$. Thus, $c_5 = 3c_1$. Now, since $c_3 = 5c_1$, it follows that $c_6 = 2c_3 = 10c_1$ by Claim 1, so $f(c_1) + f(c_5) = f(c_1) + f(3c_1) = f(10c_1) = f(c_6)$, but $\{c_1, 3c_1, 10c_1\}$ is sum-free, a contradiction. Thus, $c_3 = 3c_1$.

Now, assume that for some $k \geq 3$, we have that for all $1 \leq i \leq k$ that $c_i = ic_1$. Then, $f(c_{k+1}) = f(c_1) + f(c_k) = f(c_1) + f(kc_1)$, so $\{c_1, kc_1, c_{k+1}\}$ is not sum-free. Since $k \geq 3$, $\{c_1, kc_1\}$ is sum-free, so $c_{k+1} \in \{\frac{c_1}{2}, 2c_1, \frac{kc_1}{2}, 2kc_1, kc_1 - c_1, kc_1 + c_1\} = \left\{\frac{1}{2}c_1, 2c_1, \frac{k}{2}c_1, (k-1)c_1, (k+1)c_1, 2kc_1\right\}$. If $c_{k+1} = \frac{1}{2}c_1$, then $f(c_1) = f(2c_{k+1}) = 2f(c_{k+1}) = 2k+2$, absurd. If $c_{k+1} = 2c_1$, then $f(c_{k+1}) = f(2c_1) = 2f(c_1) = 2$, absurd. If $c_{k+1} = \frac{k}{2}c_1$, then $k = f(c_k) = f(kc_1) = f(2c_{k+1}) = 2f(c_{k+1}) = 2k+2$, absurd. If $c_{k+1} = 2kc_1$, then $f(c_{k+1}) = f(2kc_1) = 2f(kc_1) = 2f(c_k) = 2k$, absurd. If $c_{k+1} = (k-1)c_1$, $c_{k+1} = (k-1)c_1 = c_{k-1}$, absurd. Thus, $c_{k+1} = (k+1)c_1$, as desired, so our inductive step is complete and so we are done.

5. Conclusion: Since $f$ is surjective, there exists such a sequence $\{c_i\}$, so fix such a sequence. If there existed a $d_1 \neq c_1$ for which $f(d_1) = 1$, then $d_1, c_2, c_3, \cdots$ is also a valid sequence, so $c_2 = 2d_1$. But $c_2 = 2c_1$, so $c_1 = d_1$, a contradiction. So $f$ is injective at $1$. Now, $f$ is injective at any $j > 1$ since if there existed a $d_j \neq c_j$ for which $f(d_j) = j$, then $c_1, c_2, \cdots, c_{j-1}, d_j, c_{j+1}, \cdots$ is also a valid sequence, so $d_j = jc_1$. But $c_j = jc_1$ as well, so $c_j = d_j$, a contradiction. Thus, $f$ is bijective, so there exists an $i$ for which $c_i = 1$. Then, $1 = ic_1$, so $i \mid 1$, implying $i = 1$. Thus, $c_1 = 1$, so $c_i = i$, and so $f(i) = i$ for all $i \geq 1$, as desired.

$\blacksquare$

The final answer is $\boxed{ f(x) = x }$