How to prove that the sum and product of two algebraic numbers is algebraic? Suppose $E/F$ is a field extension and $\alpha, \beta \in E$ are algebraic over $F$. Then it is not too hard to see that when $\alpha$ is nonzero, $1/\alpha$ is also algebraic. If $a_0 + a_1\alpha + \cdots + a_n \alpha^n = 0$, then dividing by $\alpha^{n}$ gives $$a_0\frac{1}{\alpha^n} + a_1\frac{1}{\alpha^{n-1}} + \cdots + a_n = 0.$$

Is there a similar elementary way to show that $\alpha + \beta$ and $\alpha \beta$ are also algebraic (i.e. finding an explicit formula for a polynomial that has $\alpha + \beta$ or $\alpha\beta$ as its root)? 

The only proof I know for this fact is the one where you show that $F(\alpha, \beta) / F$ is a finite field extension and thus an algebraic extension.

Let $\alpha$ have minimal polynomial $p(x)$ and let $\beta$ have minimal polynomial $q(x)$. Then $V = F[x, y]/(p(x), q(y))$ is a finite-dimensional vector space over $F$ of dimension $\deg p \deg q$ (it is not necessarily the same dimension as $F(\alpha, \beta)$, for example when $\alpha = \beta$); moreover, it has an explicit basis
$$x^i y^j : 0 \le i < \deg p, 0 \le j < \deg q.$$
$xy$ and $x + y$ act by left multiplication on $V$ and one can write down explicit matrices for this action in the basis above in terms of the coefficients of $p$ and $q$. Now apply the Cayley-Hamilton theorem. 
This argument proves the stronger result that if $F$ is the fraction field of some domain $D$ and $\alpha, \beta$ are integral over $D$ (hence $p, q$ are monic with coefficients in $D$) then so are $\alpha \beta, \alpha + \beta$. 
