The following algorithm suggests itself. Let $\alpha$ be an unknown root of a polynomial $f(x)$ of degree $n$, and $\beta$ be an unknown root of a polynomial $g(x)$ of degree $m$.
Using the equation $f(\alpha)=0$ allows us to rewrite any power $\alpha^k$ as
a linear combination of $1,\alpha,\alpha^2,\ldots,\alpha^{n-1}$. Similarly using the equation $g(\beta)=0$ allows us to rewrite any power $\beta^\ell$ as a linear
combination of $1,\beta,\beta^2,\ldots,\beta^{m-1}$. Putting these two pieces together shows that we can write any monomial $\alpha^k\beta^\ell$ as a linear combination of the $mn$ quantities $\alpha^i\beta^j, 0\le i<n, 0\le j<m.$
Denote $c_k=\alpha^i\beta^j$, where $k=mi+j$ ranges from $0$ to $mn-1$.
Next let us use the expansions of $(\alpha\beta)^t$, $0\le t\le mn$ in terms of the $c_k$:s. Let these be
$$
(\alpha\beta)^t=\sum_{k=0}^{mn-1}a_{kt}c_k.
$$
Here the coefficients $a_{tk}$ are integers. We seek to find integers $x_t,t=0,1,\ldots,mn$, such that
$$
\sum_{t=0}^{mn}x_t(\alpha\beta)^t=0.\qquad(*)
$$
Let us substitute our formula for the power $(\alpha\beta)^t$ above. The equation $(*)$ becomes
$$
0=\sum_t\sum_kx_ta_{tk}c_k=\sum_k\left(\sum_t x_ta_{kt}\right)c_k.
$$
This will be trivially true, if the coefficients of all $c_k$:s vanish, that is, is the equation
$$
\sum_{t=0}^{mn}a_{kt}x_t=0 \qquad(**)
$$
holds for all $k=0,1,2,\ldots,mn-1$. Here there are $mn$ linear homogeneous equations on the $mn+1$ unknowns $x_t$. Therefore linear algebra says that we are guaranteed to succeed in the sense that there exists a non-trivial vector
$(x_0,x_1,\ldots,x_{mn})$ of rational numbers that is a solution of $(**)$. Furthermore, by multiplying with the least common multiple of the denominators,
we can make all the $x_t$:s integers.
The polynomial
$$
F(x)=\sum_{t=0}^{mn}x_tx^t
$$
is then an answer.
Let's do your example of $f(x)=x^2-1$ and $g(x)=x^2+3x+2$. Here $f(\alpha)=0$
tells us that $\alpha^2=1$. Similarly $g(\beta)=0$ tells us that $\beta^2=-3\beta-2$. This is all we need from the polynomials. Here $c_0=1$, $c_1=\beta$,
$c_2=\alpha$ and $c_3=\alpha\beta$. Let us write the power $(\alpha\beta)^t$,
$t=0,1,2,3,4$ in terms of the $c_k$:s.
$$
(\alpha\beta)^0=1=c_0.
$$
$$
(\alpha\beta)^1=\alpha\beta=c_3.
$$
$$
(\alpha\beta)^2=\alpha^2\beta^2=1\cdot(-3\beta-2)=-2c_0-3c_1.
$$
$$
(\alpha\beta)^3=\alpha\beta(-3\beta-2)=\alpha(-3\beta^2-2\beta)=\alpha(9\beta+6-2\beta)=\alpha(7\beta+6)=6c_2+7c_3.
$$
$$
(\alpha\beta)^4=(-3\beta-2)^2=9\beta^2+12\beta+4=(-27\beta-18)+12\beta+4=-14-15\beta=-14c_0-15c_1.
$$
We are thus looking for solutions of the homogeneous linear system
$$
\left(\begin{array}{ccccc}
1&0&-2&0&-14\\
0&0&-3&0&-15\\
0&0&0&6&0\\
0&1&0&7&0
\end{array}\right)
\left(\begin{array}{c}x_0\\x_1\\x_2\\x_3\\x_4\end{array}\right)=0.
$$
Let us look for a solution with $x_4=1$ (if the polynomials you started with
are monic, then this always works AFAICT). The third equation tells us that we should have $x_3=0$. The second equation allows us to solve that $x_2=-5$. The last equation and our knowledge of $x_3$ tells that $x_1=0$. The first equation
then tells that $x_0=4.$ This gives us the output
$$
x^4-5x^2+4.
$$
The possibilities for $\alpha$ are $\pm1$ and the possibilities for $\beta$ are $-1$ and $-2$. We can easily check that all the possible four products $\pm1$ and $\pm2$ are zeros of this quartic polynomial.
My solution to the linear system was ad hoc, but there are known algorithms for that: elimination, an explicit solution in terms of minors,...
There is no nice formula to get the roots of $P+Q$ from the roots of $P$ and of $Q$. For example, the roots of $x^5$ and $2x+1$ are easy to find, but
the sum of these polynomials is $x^5 + 2 x + 1$, an irreducible quintic whose roots can't be expressed in radicals.
EDIT: In the case of a polynomial with three or four terms, we can express it as the sum of two polynomials with one or two terms, and then there is only trivial "additional information about the polynomials". Any "analytic method" is going to have to find roots of polynomials with three or four terms. And then once you have those roots, you can apply your "method" again to find roots of polynomials with five to eight terms...
Best Answer
The sum of the coefficients is just the value of the polynomial in $1$, hence if the polynomial is monic with roots in $n_1,\ldots,n_k$, then $$p(1) = (1-n_1)\cdot\ldots\cdot(1-n_k).$$