If I undestood the question correctly we have: \[ x_1+x_2x_3x_4=2 \] and so on.So we have: \[ x_1^2-2x_1+x_1x_2x_3x_4=0 \] And for that if we make \[ k=x_1x_2x_3x_4 \] a constant then \[ x_i \] with \[ i=1,2,3,4 \] can only equal one of the 2 rooths of the equation: \[ x^2-2x+k=0 \]. And for that we have only 3 possibilitys: \[ x_1=x_2=x_3=x_4; or; x_1=x_2, and, x_3=x_4; or; x_1=x_2=x_3, and, x_4 \]. In the first possibility we have the equation \[ x+x^3-2=0 \] and the only real rooth to this equation is x=1 and we have the set \[ {1,1,1,1} \] If \[ x_1=x_2=a;and;x_3=x_4=b \] we have that the rooths of \[ x^2-2x+a^2b^2=0 \] are a;b so we have \[ ab=(a^2b^2)/1 \] and ab=0 or 1. If ab=0 then a and b are the rooths of \[ x^2-2x=0 \] so (a,b)=(0,2) but {2,2,0,0} is not a set with the propriety. If ab=1 then a and b are the rooths of \[ x^2-2x+1=0 \] so (a,b)=(1,1) and we have the set {1,1,1,1} again. Now we only have to check the case that \[ x_1=x_2=x_3=a;and;x_4=b \]. We have that a and b are the rooths of \[ x^2-2x+a^3b=0 \] so a+b=2 and ab=a^3b or \[ a^2=1 \] and we have the sets {1,1,1,1} or {-1,-1,-1,3}. or ab=0. If ab=0 then a and b are rooths of \[ x^2-2x=0 \] so (a,b)=(0,2) but neither {2,2,2,0} or {0,0,0,2} are good sets. So the sets are {1,1,1,1} and {3,-1,-1,-1}. If there are any mistakes in this solution please tell me.

I think you could speed up the reasoning if you consider two relations and then substract them. For example $(x_1-x_2)(1-x_3x_4)=0$.

$a=x_{1}x_{2}x_{3},b=x_{1}x_{2}x_{4},c=x_{1}x_{3}x_{4},d=x_{2}x_{3}x_{4}$ so $x_{1}+d=2 $ and so on. from equation 2 we get that $\frac{2x_{2}-x_{2}^{2}}{d}=x_{1}$ from equation 1 we know that $x_{1}=2-d$ from these two equation we get $2d-d^{2}=2x_{2}-x_{2}^{2}$ so with same way we get $2c-c^{2}=2x_{2}-x_{2}^{2}$ so $(d-c)(2-d-c)=0$ then $d=c$or$2=d+c$ $2=d+c$ so $x_{1}+x_{2}=2$so $x_{3}x_{4}=1$ multiply $x_{3}$ with equation 3 and multiply $x_{4}$ with equation 4 we get $x_{3}=x_{4}$ or $x_{3}=-x_{4}$ with equation 1 and 3 we get $ (x_{1}-x_{3})(1-x_{3}x_{2})=0 $so then is obvious $d=c$ it's some how similar. solution is $(x_{1},x_{2},x_{3},x_{4})=(1,1,1,1),(-1,-1,-1,3)$

Observe that subtracting the equations $x_1+x_2x_3x_4=2~(1)$ and $x_2+x_3x_4x_1=2~(2)$ yields $x_1-x_2-x_3x_4x_1+x_3x_4x_2=0$, which rearranges as $$(x_1-x_2)-x_3x_4(x_1-x_2)=(x_1-x_2)(1-x_3x_4)=0\implies x_1=x_2\text{ or }x_3x_4=1.$$In other words, subtracting equations $(1)$ and $(2)$ implies that $x_1=x_2$ or $x_3x_4=1$, and generally subtracting equations $(a)$ and $(b)$, where $1\leq a\neq b\leq 4$, implies that $x_a=x_b$ or $x_cx_d=1$, where $\{a,b,c,d\}=\{1,2,3,4\}$. It follows that $x_a=x_b$ or $x_cx_d=1$ for all such $a,b$. Now, we consider how many of the $x_k$ are equal. If none of them are equal, we have that $x_cx_d=1$ for all $c,d$, which yields $(1,1,1,1)$, a contradiction. Similarly, if only $2$ of them are equal, we have that $x_cx_d=1$ for all $c,d$ except for one pair, but regardless we again have $(1,1,1,1)$, impossible. Furthermore, if $2$ of them are equal and the other $2$ are equal, WLOG let $x_1=x_2$ and $x_3=x_4$; then $x_2x_3=1$ since $x_1\neq x_4$ so $x_3=x_4=\tfrac{1}{x_2}$, but plugging into $x_1+x_2x_3x_4=2$ gives $x_1+\tfrac{1}{x_1}=2$, whose only real solution is $x_1=1$. Consequently we have $(1,1,1,1)$ again, impossible. Hence, we have shown that either $3$ of them are equal and the other is distinct, or all $4$ of them are equal, the latter of which yields the solution $\boxed{(1,1,1,1)}$. In the former case, WLOG let $x_1=x_2=x_3$. Then since $x_1\neq x_4$, $x_2x_3=x_2^2=1$, so $x_1=x_2=x_3=\pm1$. Having $x_1=x_2=x_3=1$ yields $(1,1,1,1)$ once again, which is impossible, so we take $x_1=x_2=x_3=-1$, which yields $\boxed{(-1,-1,-1,3)\text{ and permutations}}$ upon substituting back into $x_1+x_2x_3x_4=2$. All cases have been exhausted, so we're done.

Subtracting the first $2$ equations gives us $(x_1-x_2)(1-x_3 \cdot x_4)=0$, so we have $x_1=x_2$ or $x_3 \cdot x_4 =1.$ Now we proceed with bashy casework. After doing this casework we get $(1,1,1,1)$ and $(-1,-1,-1,3)$ and permutations. $\square$