If $x=1$ then $y=z$ and $y+z=1$ hence $x=y=z=1$ in which case we have $s=3$. Similarly if $y=1$ or $z=1$ we get $s=3$. From now on we consider $x,y,z \neq 1$. This also gives $x \neq y \neq z \neq x$. Solving the first expression $x=\frac{2z-y}{z}$ then plugging this in to the second two gives: $$y+\frac{z^2}{2z-y}=2 \Rightarrow (2z-y)y+z^2=2(2z-y)$$$$z+\frac{2z-y}{yz}=2 \Rightarrow yz^2+2z-y=2yz \Rightarrow y=-\frac{2z}{z^2-2z-1}$$Plugging the latter in to the former and simplifying gives: $$\frac{z^2 (z^4-8z^3+14z^2-7)}{(z^2-2z-1)^2}=0 \Rightarrow z^4-8z^3+14z^2-7=0$$As we can't have $z=0$ this reduces to: $$0=z^4-8z^3+14z^2-7=(z-1)(z^3-7z^2+7z+7) \Rightarrow z^3-7z^2+7z+7=0$$as $z \neq 1$. But by cyclic symmetry, $x,y$ also satisfy this cubic and as $x,y,z$ are distinct they are precisely the $3$ roots of the above cubic (if the expression holds). In this case by Vieta formula $s=7$. We've now covered all cases so $s=3$ or $s=7$

Note that we can derive that $(x,y,z)=(1,1,1)$ or a cyclic permutation of $\left(4\cos\frac{2\pi}{7}+3,4\cos\frac{4\pi}{7}+3,4\cos\frac{6\pi}{7}+3\right)$. The former gives $S=3$ and the latter gives $S=7$.

Another soltution: In the beginning, we prove that $x+y+z=xy+yz+zx=s$ just like above. Then, we transform the $3$ equations and multiply: $$y=z(2-x)$$$$z=x(2-y)$$$$x=y(2-z)$$$$(2-x)(2-y)(2-z)=1 \Longleftrightarrow$$$$8-4(x+y+z)+2(xy+yz+zx)-xyz=1$$$$\Longleftrightarrow xyz=7-2s$$Note that $x, y, z \neq{0}$, so $x, y, z \neq{2}$. Now, multiplying the equations as rewritten by @above with $y$, $z$, $x$ respectively and summing gives: $$xyz+y^2=2zy$$$$xyz+z^2=2xz$$$$xyz+x^2=2yx$$$$3xyz+x^2+y^2+z^2=2(xy+yz+zx) \Longleftrightarrow$$$$3(7-2s)+(x+y+z)^2-2(xy+yz+zx)=2(xy+yz+zx) \Longleftrightarrow$$$$21-6s+s^2-2s=2s \Longleftrightarrow$$$$s^2-10s+21=0 \Longleftrightarrow$$$s=3$ or $s=7$, as desired.

Here is my solution: Ideally, we wish to show that $T\coloneqq s^2-10s+21=(s-3)(s-7)=0$. Expanded, this is \[T=x^2+y^2+z^2+2(xy+yz+zx)-10(x+y+z)+21\]Multiplying each equation by its denominator, we get $zx+y=2z$ and its symmetric expressions. Adding them up gives \[xy+yz+zx=x+y+z\]Multiplying all together yields \[x^2+y^2+z^2+\frac{zx}{y}+\frac{xy}{z}+\frac{yz}{x}+xyz=7\]Note that $\frac xy=2-z$ (also symmetricaly for the others). Plugging this in and canceling, we see \[2(x+y+z)+xyz=7.\]We can get a hold of $xyz$ by multipying the first expressions by $yz$ to get $xyz+y^2=2yz$. Adding this symmetrically up gives us \[3xyz+(x^2+y^2+z^2)=2(xy+yz+zx)\Longleftrightarrow xyz=\frac{2(xy+yz+zx)-(x^2+y^2+z^2)}{3}\]Plugging this into the previous expression and multiplying by $3$ gives \[6(x+y+z)+2(xy+yz+zx)-(x^2+y^2+z^2)=21\]Multiplying by $-1$ yields \[x^2+y^2+z^2-6(x+y+z)-2(xy+yz+zx)+21=0\]Finally, adding and subtracting $2(xy+yz+zx)$ on the LHS and using $xy+yz+zx=x+y+z$ gives \[(x+y+z)^2-10(x+y+z)+21=0,\]as desired.