Solution:
We have $x^{2}+y^{2}+2 x y=1$. Define $a=2 x y$ and $b=x^{2}+y^{2}$ for convenience. Then $a+b=1$ and $b-a=x^{2}+y^{2}-2 x y=(x-y)^{2}=2 b-1$ so that $x, y=\frac{\sqrt{2 b-1} \pm 1}{2}$. Then
$$\begin{aligned} x^{20}+y^{20} & =\left(\frac{\sqrt{2 b-1}+1}{2}\right)^{20}+\left(\frac{\sqrt{2 b-1}-1}{2}\right)^{20} \\ & =\frac{1}{2^{20}}\left[(\sqrt{2 b-1}+1)^{20}+(\sqrt{2 b-1}-1)^{20}\right] \\ & =\frac{2}{2^{20}}\left[(\sqrt{2 b-1})^{20}+\binom{20}{2}(\sqrt{2 b-1})^{18}+\binom{20}{4}(\sqrt{2 b-1})^{16}+\ldots\right] \\ & =\frac{2}{2^{20}}\left[(2 b-1)^{10}+\binom{20}{2}(2 b-1)^{9}+\binom{20}{4}(2 b-1)^{8}+\ldots\right] \\ & =20 \end{aligned}$$
We want to find the sum of distinct roots of the above polynomial in $b$; we first prove that the original polynomial is square-free. The conditions $x+y=1$ and $x^{20}+y^{20}=20$ imply that $x^{20}+(1-x)^{20}-20=0$; let $p(x)=x^{20}+(1-x)^{20}-20$. $p$ is square-free if and only if $\gcd\left(p, p'\right)=c$ for some constant $c$:
$$\begin{aligned} \gcd\left(p, p'\right) & =\gcd\left(x^{20}+(1-x)^{20}-20,20\left(x^{19}-(1-x)^{19}\right)\right) \\ & =\gcd\left(x^{20}-x(1-x)^{19}+(1-x)^{19}-20,20\left(x^{19}-(1-x)^{19}\right)\right) \\ & =\gcd\left((1-x)^{19}-20, x^{19}-(1-x)^{19}\right) \\ & =\gcd\left((1-x)^{19}-20, x^{19}-20\right) \end{aligned}$$
The roots of $x^{19}-20$ are $\sqrt[19]{20^{k}} \exp \left(\frac{2 \pi i k}{19}\right)$ for some $k=0,1, \ldots, 18$; the roots of $(1-x)^{19}-20$ are $1-\sqrt[19]{20^{k}} \exp \left(\frac{2 \pi i k}{19}\right)$ for some $k=0,1, \ldots, 18$. If $x^{19}-20$ and $(1-x)^{19}-20$ share a common root, then there exist integers $m$, $n$ such that $\sqrt[19]{20^{m}} \exp \left(\frac{2 \pi i m}{19}\right)=1-\sqrt[19]{20^{n}} \exp \left(\frac{2 \pi i n}{19}\right)$; since the imaginary parts of both sides must be the same, we have $m=n$ and $\sqrt[19]{20^{m}} \exp \left(\frac{2 \pi i m}{19}\right)=\frac{1}{2} \Longrightarrow 20^{m}=\frac{1}{2^{19}}$, a contradiction. Thus we have proved that the polynomial in $x$ has no double roots. Since for each $b$ there exists a unique pair $(x, y)$ (up to permutations) that satisfies $x^{2}+y^{2}=b$ and $(x+y)^{2}=1$, the polynomial in $b$ has no double roots.
Let the coefficient of $b^{n}$ in the above equation be $\left[b^{n}\right]$. By Vieta's Formulas, the sum of all possible values of $b=x^{2}+y^{2}$ is equal to $-\frac{\left[b^{9}\right]}{\left[b^{10}\right]}$. $\left[b^{10}\right]=\frac{2}{2^{20}}\left(2^{10}\right)$ and $\left[b^{9}\right]=\frac{2}{2^{20}}\left(-\binom{10}{1} 2^{9}+\binom{20}{2} 2^{9}\right)$; thus $-\frac{\left[b^{9}\right]}{\left[b^{10}\right]}=-\frac{\binom{10}{1} 2^{9}-\binom{20}{2} 2^{9}}{2^{10}}=-90$.