[EXPANDED]

**PART 1** (also done by Peter)

If **$x,y$ are coprime and have opposite sign**, there is a singular symmetric matrix with 1 on the diagonal and
only $x$ and $y$ off the diagonal.

Say $x<0,y>0$. Since $x$ and $y$ are coprime, there are positive integers $n_1,n_2$ such that $n_1 x + n_2 y = -1$.
Moreover, $n_1$ and $n_2$ can be chosen to have odd sum, since at least one of $n_1 x + n_2 y = -1$ and
$(n_1+y)x + (n2-x)y=-1$ has that property.

Representative examples: $5(-3)+2(7)=-1$, $11(-3)+8(4)=-1$, $4(-4)+5(3)=-1$.

Now make a symmetric matrix of order $n_1+n_2+1$ such that each row has, apart from its diagonal 1, $n_1$ values equal to $x$ and $n_2$ values equal to $y$. This is always possible as a complete graph of even order can be decomposed into hamiltonian cycles and one perfect matching.

The matrix has all row sums equal to 0 so it is singular.

Here is the case $5(-3)+2(7)=-1$
$$\pmatrix{
1 & -3 & -3 & 7 & -3 & 7 & -3 & -3
\\
-3 & 1 & -3 & -3 & 7 & -3 & 7 & -3
\\
-3 & -3 & 1 & -3 & -3 & 7 & -3 & 7
\\
7 & -3 & -3 & 1 & -3 & -3 & 7 & -3
\\
-3 & 7 & -3 & -3 & 1 & -3 & -3 & 7
\\
7 & -3 & 7 & -3 & -3 & 1 & -3 & -3
\\
-3 & 7 & -3 & 7 & -3 & -3 & 1 & -3
\\
-3 & -3 & 7 & -3 & 7 & -3 & -3 & 1
}$$

**PART 2**

Next, we give a general solution for **coprime $x,y$,
where $x\ge 2,y\ge x+2$**.
Find integer $n_2,k$ such that $n_2y+1=kx$; this is possible because $x,y$ are coprime. Now define $n_1=(k-n_2-1)x-k$. It is possible to choose $n_2,k$ such that $n_1,n_2$ are positive and have opposite parity (if necessary add $ty$ to $k$ and $tx$ to $n_2$ for some $t\ge 1$). Also define $n_3=x-1$. Now consider the matrix
$$\pmatrix{A&B\\B^T&C}$$ defined like this:
$A$ is square of order $n_1+n_2+1$, with each row having one 1 (on the diagonal), $n_1$ of $x$ and $n_2$ of $y$. $C$ is square of order $n_3$. Apart from the diagonal of $C$, which of course is 1, $B$ and $C$ are filled with $x$.
This matrix is singular because the vector with first $n_1+n_2+1$ entries equal to $n_3x$ and last $n_3$ entries equal to $-(n_1x+n_2y+1)$ is an eigenvector for eigenvalue 0.

**PART 3**

Next, we give a general solution for **coprime $x,y$,
where $x\le -2,y\le x-2$**.

This is nearly the same as the previous case. Find integer $n_2,k$ such that $n_2y+1=kx$. Now define $n_1=(n_2-k+1)x-k$. It is possible to choose $n_2,k$ such that $n_1,n_2$ are positive and have opposite parity (if necessary add $ty$ to $k$ and $tx$ to $n_2$ for some $t\ge 1$). Also define $n_3=-x+1$. Now construct the matrix as before.

**CONCLUSION**

In previous comments it was noted that the cases $-1\in S$, $1\in S$ and $\{k,k+1\}\subseteq S$ have simple constructions. So we have a theorem.

**THEOREM** If $x,y$ are distinct coprime integers, then there is a singular symmetric matrix with 1 on the diagonal and $x$ or $y$ in every off-diagonal position.

Note that this does not complete the solution to the original problem, as having a set $S$ with no nontrivial common factors does not mean that it contains two coprime elements. Consider $S=\{6,10,15\}$ for example.