Problem 1939

The only solution is $f(a) = 0$ for $a \notin S$ and $f(x) = x$ for $x \in S$.
Label the equations as follows.
$$f(a + x + y) + f(f(a)) + f(x) + f(y) = x + y \quad (1)$$
$$f(axy) + f(a) + f(x)f(y) = xy \tag{2}$$

Case 1. $0 \notin S$
Putting $a = 0$ in (2), we obtain
$$f(x)f(y) = xy - 2f(0) \tag{3}$$
for all $x, y \in S$. In particular, by putting $x = y$, we get
$$f(x)^2 = x^2 - 2f(0) \tag{4}$$
for all $x \in S$. Then we have
$$(x^2 - 2f(0))(y^2 - 2f(0)) = f(x)^2 f(y)^2 = (xy - 2f(0))^2,$$
which implies $2f(0)(x^2 + y^2) = 4f(0)xy$. This means
$$2f(0)(x - y)^2 = 0.$$
Since $S$ has at least two elements, we can choose distinct $x, y \in S$ to conclude
that $f(0) = 0$. Therefore, by (4),
$f(x)^2 = x^2$
for all $x \in S$. Since $f(x)f(y) = xy$ by (3), we see that the choice of the sign for $f(x)$ is independent of $x$. This means $f(x) = x$ for all $x \in S$ or $f(x) = -x$ for all $x \in S$.
Now, (2) is reduced to
$$f(axy) + f(a) = 0. \tag{5}$$

For fixed nonzero $a \notin S$, if $\frac{1}{a} \in S$, we may put $y = \frac{1}{a}$ in (5) to get $f(x) + f(a) = 0$. But this cannot be true as we can choose two different values for $x$ (and hence $f(x)$). Therefore, we must have $\frac{1}{a} \notin S$. From this, we see that $x \in S$ implies $\frac{1}{x} \in S$, since otherwise $x = (x^{-1})^{-1} \notin S$.
Now, we can put $y = \frac{1}{x}$ in (5) to obtain $2f(a) = 0$, i.e. $f(a) = 0$ for any $a \notin S$. Equation (1) becomes
$$f(a + x + y) \pm (x + y) = x + y.$$
By putting $x = y$, we get
$$f(a + 2x) \pm 2x = 2x.$$
If the negative sign is chosen, then we have $f(a + 2x) = 4x$. Since $f(a + 2x)$ can only be $-(a+2x)$ or $0$ (depending on whether $a+2x \in S$ or not), we need $a = -6x$ for any $a \notin S$ and $x \in S$ (note we cannot have $0 = 4x$). This is impossible as we can find two distinct elements in $S$. Therefore, $f(a) = 0$ for $a \notin S$ and $f(x) = x$ for $x \in S$.

Case 2. $0 \in S$
Putting $y = 0$ in (2), we obtain
$$f(0) + f(a) + f(x)f(0) = 0 \qquad (6)$$
for all $a \notin S$, $x \in S$. In particular, by putting $x = 0$, we get
$$f(a) = -f(0) - f(0)^2. \qquad (7)$$
Then equation (6) becomes
$$f(x)f(0) = f(0)^2.$$
If $f(0) \neq 0$, we need $f(x) = f(0)$ for all $x \in S$. Consider equation (1). It now becomes
$$f(a + x + y) + f(-f(0) - f(0)^2) + 2f(0) = x + y.$$
Note that the left-hand side can take at most two values (as $f(a+x+y)$ can be $-f(0) - f(0)^2$ or $f(0)$). However, as there are at least two elements in $S$, say 0 and $z \neq 0$, the right-hand side can take at least three different values, namely, 0, $z$, $2z$. This is a contradiction, and hence $f(0) = 0$. By (7), we obtain $f(a) = 0$ for $a \notin S$.
Next, we prove that $a \notin S$ implies $-a \notin S$. Indeed, suppose $-a \in S$. We put $y = -a$ in (1) to get
$$2f(x) + f(-a) = x - a. \qquad (8)$$
$$f(-a) = -a, \tag{9}$$
and hence $2f(x) = x$ for any $x \in S$ by (8). Now, by putting $x = -a$, we obtain $2f(-a) = -a$. But then we have $f(-a) = -a$ by (9). This forces $a = 0 \in S$, which is a contradiction. So the claim is true. From this, we see that $x \in S$ implies $-x \in S$. Thus, we can put $y = -x$ in (1). This gives
$$f(x) + f(-x) = 0 \tag{10}$$
for any $x \in S$.
Similarly, we shall prove that $z, w \in S$ implies $z+w \in S$. Suppose on the contrary that $z+w \notin S$. Note that $-w \in S$ from above. So we can put $a = z+w$ and $x = 0$, $y = -w$ in (1) to obtain
$$f(z) + f(-w) = -w.$$
Using (10), we get
$$f(z) - f(w) = -w.$$
By symmetry, we also have
$$f(w) - f(z) = -z.$$
Adding these, we obtain $z+w=0 \in S$, which is a contradiction.
It is now clear that $a+x+y \notin S$ for any $a \notin S$ and $x, y \in S$. Otherwise, if $a+x+y \in S$, since $-x-y = (-x)+(-y) \in S$, we have
$$a = (a + x + y) + (-x - y) \in S.$$
It follows that (1) is reduced to $f(x)+f(y)=x+y$. Putting $x=y$, we get $f(x)=x$ for any $x \in S$.
In any case, $f(a) = 0$ for $a \notin S$ and $f(x) = x$ for $x \in S$ is the only possible function. One can check that this is indeed a solution if and only if $a+x+y, axy \notin S - \{0\}$ for any $a \notin S$ and $x, y \in S$.