The complete set of solutions consists of
$$(1,0), (0,1),$$
$$(a,x), (x,a), (1/a,-x/a), (-x/a,1/a), (1/x,-a/x), (-a/x,1/x),$$
$$(b,y), (y,b), (1/b,-y/b), (-y/b,1/b), (1/y,-b/y), (-b/y,1/y).$$

Let $C$ be the affine curve $X^5+Y^5=1$ over $\mathbf{Q}$. Because your field $K:=\mathbf{Q}(a,b)[x,y]$ is the function field of the $\mathbf{Q}$-variety $C \times C$, the set $C(K)$ is in bijection with the set of rational maps $C \times C \to C$. So the only fact needed beyond what was in FC's answer to your earlier question is the geometric fact that every non-constant rational map $C \times C \to C$ is a composition consisting of the first or second projection $C \times C \to C$ followed by a birational automorphism of $C$. More generally,

If $X,Y,Z$ are curves over a field $k$, and the genus of $Z$ is at least $2$, then every rational map $X \times Y \to Z$ factors through the first or second projection.

**Proof:** Extend the rational map to a rational map $X \times Y \to Z \times Y$ by using the projection $X \times Y \to Y$ as the second coordinate map. Restrict this to the fibers above the generic point of $Y$ to get a rational map $X_L \to Z_L$, where $L$ is the function field of $Y$. If this is constant, i.e., factors through the structure map to $\operatorname{Spec L}$, then the original map factors through the projection to $Y$. On the other hand, the genus hypothesis implies that, up to powers of Frobenius in characteristic $p$, there are only finitely many non-constant rational maps from $X$ to $Z$ over any field, and hence finitely many of bounded degree in any characteristic, so there are no algebraic families of such maps, so the rational map $X_L \to Z_L$ must be the base extension of a rational map $X \to Z$, which means that the original rational map factors through the projection to $X$. $\square$

@Bakh: Both FC and I used the general observation that if $X$ and $Y$ are curves over a field $k$, and $K$ is the function field of $Y$, then $X(K)$ is in bijection with the set of rational maps from $Y$ to $X$. If you have not studied enough algebraic geometry yet to understand this, the following example may be helpful:

Consider the case where $X$ is the plane curve $f(x,y)=0$ and $Y$ is the affine line $\mathbf{A}^1$, so $K=k(t)$. To give a point in $X(K)$ is to give rational functions $r(t)$ and $s(t)$ such that $f(r(t),s(t))=0$ identically. If we substitute a particular element of $k$ for $t$, then $(r(t),s(t))$ specializes to a point on $X(k)$, and hence we get a map $\mathbf{A}^1(k) \to X(k)$ except that we must avoid the finitely many poles of $r(t)$ and $s(t)$. The same holds for points over any field extension of $k$, and in fact we get a rational map $\mathbf{A}^1 \to X$.

More scheme-theoretically, one can say that an element of $X(K)$ is a $k$-morphism from the generic point of $Y$ to $X$, and such a morphism spreads out to a morphism from some Zariski dense open subscheme of $Y$ to $X$, i.e., to a rational map from $Y$ to $X$.