Yes.

Two possible examples are
$$(N^2 - 3N + 1,\ N^2 - N + 1,\ -3N^2 + 3N - 1,\ N^2 + N - 1)$$
and
$$(N^3 - N^2 + 1,\ N^3 - 3N^2 + 2N - 1,\ -N^3 + N^2 - 2N + 1,\ -N^3 + N^2 + 2N - 1),$$
where $N$ is an integer greater than $10^6$.

First solution. For example, the numbers
$$x = N^2 - 3N + 1,\ y = N^2 - N + 1,\ z = -3N^2 + 3N - 1,\ t = N^2 + N - 1,$$
where $N$ is a natural number greater than a million.

It is easy to see that these numbers are also greater than a million in absolute value and pairwise distinct. Their pairwise sums are
$$x + y = 2(N - 1)^2,\ x + z = -2N^2,\ y + z = -2N(N - 1),\ z + t = -2(N - 1)^2,\ y + t = 2N^2,\ x + t = 2N(N - 1),$$
with the products
$$(x + y)(x + z) = (y + t)(z + t) = (x + t)(y + z) = -4N^2(N - 1)^2.$$

It remains to check that the numbers $x, y, z, t$ are coprime as a set. If they have a common natural divisor $d$, then $d$ also divides $x + y = 2(N - 1)^2$ and $y + t = 2N^2$; thus, $d \le \gcd(2N^2, 2(N - 1)^2) = 2$. The case $d = 2$ is impossible since $y = N(N - 1) + 1$ is odd. Therefore, $d = 1$, as required.

Second solution. Another example is
$$x = N^3 - N^2 + 1,\ y = N^3 - 3N^2 + 2N - 1,\ z = -N^3 + N^2 - 2N + 1,\ t = -N^3 + N^2 + 2N - 1,$$
where $N$ is a natural number greater than a million. Again, these numbers are greater than a million in absolute value and pairwise distinct. Their pairwise sums are
$$x + y = 2N(N - 1)^2,\ x + z = -2(N - 1),\ y + z = -2N^2,\ z + t = -2N^2(N - 1),\ y + t = -2(N - 1)^2,\ x + t = 2N,$$
and they can be grouped into three pairs with equal products:
$$(x + y)(x + t) = (x + z)(z + t) = (y + z)(y + t) = 4N^2(N - 1)^2.$$

The coprimality of $x, y, z, t$ is checked analogously to the previous solution.