Let $p(x)=\prod_i (x-\alpha_i)$. Then every symmetric function in the $\alpha_i$ is a polynomial in the elementary symmetric functions, which are $\pm$ the coefficients of $p$. Hence, it must lie in the base field.
If $q(x)=\prod_j (x-\beta_j)$, then $\prod_{i,j} (x-\alpha_i - \beta_j)$ can be considered as a polynomial in the $x,\alpha_i,\beta_j$. It is symmetric in the $\alpha_i$, so that it is a polynomial in $x,\beta_j$ with coefficients in the base field. But it is also symmetric in the $\beta_j$, hence is a polynomial in $x$ with coefficients in the base field.
The same argument works for $\prod_{i,j} (x-\alpha_i \beta_j)$.
When $p,q$ are irreducible, it is not true in general that $\prod_{i,j} (x-\alpha_i-\beta_j)$ and $\prod_{i,j} (x-\alpha_i \beta_j)$ are irreducible. This can already been seen from their degrees, which are too large, for example if $\alpha_1=-\beta_1$ or $\alpha_1=\beta_1^{-1}$.
Actually we don't need fields here. The same argument can be used to show that the sum and the product of two integral elements is again integral, for arbitrary ring extensions.
PS: The proof above is really constructive. Let me illustrate this for two quadratic polynomials $x^2-px+q=(x-\alpha_1)(x-\alpha_2)$ and $x^2-rx+s=(x-\beta_1)(x-\beta_2)$, so that $p=\alpha_1+\alpha_2$, $q=\alpha_1 \alpha_2$ and $r=\beta_1+\beta_2$, $s=\beta_1 \beta_2$. Then we compute:
$(x - a_1 - b_1) (x - a_1 - b_2) (x - a_2 - b_1) (x - a_2 - b_2)$
$=x^4 - 2 b_1 x^3 - 2 b_2 x^3 - 2 x^3 (\mathbf{a_1 + a_2}) + b_1^2 x^2 + 4 b_1 b_2 x^2 + b_1 x^2 (\mathbf{3 a_1 + 3 a_2}) + b_2^2 x^2 + b_2 x^2 (\mathbf{3 a_1 + 3 a_2}) + x^2 (\mathbf{a_1^2 + 4 a_1 a_2 + a_2^2}) - 2 b_1^2 b_2 x - b_1^2 x (\mathbf{a_1 + a_2}) - 2 b_1 b_2^2 x - 4 b_1 b_2 x (\mathbf{a_1 + a_2}) - b_1 x (\mathbf{a_1^2 + 4 a_1 a_2 + a_2^2}) - b_2^2 x (\mathbf{a_1 + a_2}) - b_2 x (\mathbf{a_1^2 + 4 a_1 a_2 + a_2^2}) - 2 \mathbf{a_1 a_2} x (\mathbf{a_1 + a_2}) + b_1^2 b_2^2 + b_1^2 b_2 (\mathbf{a_1 + a_2}) + \mathbf{a_1 a_2} b_1^2 + b_1 b_2^2 (\mathbf{a_1 + a_2}) + b_1 b_2 (\mathbf{a_1 + a_2})^2 + \mathbf{a_1 a_2} b_1 (\mathbf{a_1 + a_2}) + \mathbf{a_1 a_2} b_2^2 + \mathbf{a_1 a_2} b_2 (\mathbf{a_1 + a_2}) + \mathbf{a_1^2 a_2^2}$
$=x^4 - 2 b_1 x^3 - 2 b_2 x^3 - 2 x^3 p + b_1^2 x^2 + 4 b_1 b_2 x^2 + b_1 x^2 (3 p) + b_2^2 x^2 + b_2 x^2 (3 p) + x^2 (p^2 + 2q) - 2 b_1^2 b_2 x - b_1^2 x p - 2 b_1 b_2^2 x - 4 b_1 b_2 x p - b_1 x (p^2 + 2q) - b_2^2 x p - b_2 x (p^2 + 2q) - 2 q x p + b_1^2 b_2^2 + b_1^2 b_2 p + q b_1^2 + b_1 b_2^2 p + b_1 b_2 p^2 + q b_1 p + q b_2^2 + q b_2 p + q^2$
$=x^4 - 2 p x^3 - 2 x^3 (\mathbf{b_1 + b_2}) + p^2 x^2 + p x^2 (\mathbf{3 b_1 + 3 b_2}) + 2 q x^2 + x^2 (\mathbf{b_1^2 + 4 b_1 b_2 + b_2^2}) - p^2 x (\mathbf{b_1 + b_2}) - 2 p q x - p x (\mathbf{b_1^2 + 4 b_1 b_2 + b_2^2}) - 2 q x (\mathbf{b_1 + b_2}) - 2 \mathbf{b_1 b_2} x (\mathbf{b_1 + b_2}) + \mathbf{b_1 b_2} p^2 + p q (\mathbf{b_1 + b_2}) + \mathbf{b_1 b_2} p (\mathbf{b_1 + b_2}) + q^2 + q (\mathbf{b_1^2 + b_2^2}) + \mathbf{b_1^2 b_2^2}$
$=x^4 - 2 p x^3 - 2 x^3 (r) + p^2 x^2 + p x^2 (3 r) + 2 q x^2 + x^2 (r^2+2s) - p^2 x (r) - 2 p q x - p x (r^2+2s) - 2 q x (r) - 2 b_1 b_2 x (r) + s p^2 + p q (r) + s p (r) + q^2 + q (r^2-2s) + s^2$
This polynomial has coefficients in $\mathbb{Z}[p,q,r,s]$.
