The answer is $s$ composite.

Composite construction Write $s = (w+x)(y+z)$, where $w, x, y, z$ are positive integers. Let $a = wy$, $b = wz$, $c = xy$, $d = xz$. Then
$$abc + abd + acd + bcd = wxyz(w+x)(y+z)$$
so this works.

Prime proof Choose suitable $a, b, c, d$. Then
$$(a+b)(a+c)(a+d) = (abc+abd+acd+bcd) + a^2(a+b+c+d) \equiv 0 \pmod{s}.$$
Hence $s$ divides a product of positive integers less than $s$, so $s$ is composite.

Remark. Here is another proof that $s$ is composite.
Suppose that $s$ is prime. Then the polynomial $(x-a)(x-b)(x-c)(x-d) \in \mathbb{F}_s[x]$ is even, so the roots come in two opposite pairs in $\mathbb{F}_s$. Thus the sum of each pair is at least $s$, so the sum of all four is at least $2s > s$, contradiction.