Can you find an example of set $S$ with implicit representation $F(x)+G(y)=0$, where $F,G:\mathbb R \to \mathbb R$ are continuous functions such that:
$\ $ i) $S$ is a union of two disjoint endless simple curves $C_1, C_2$;
$\ $ ii) $C_1$ is not a mirror image of $C_2$.
Graph of any continuous function $f$ can be implicitly represented by $f(x)-y=0$. Union of graphs of functions $f$ and $g$ can be implicitly represented by $$ 0 = [f(x)-y][g(x)-y] = f(x)g(x) - y^2 - [f(x)+g(x)]y, $$ which however is separable into the form $F(x)+G(y)$ iff $f(x)+g(x)$ is constant, implying that the graphs of $f$ and $g$ are horizontal mirror of each other.
Another example is the equation $$ y^2 + \max\{0,1-|x+1| \} = 0, $$ which implicitly characterizes the union of two half-lines $\{(x,0): x\leq 0\}$ and $\{(x,0): x\geq 2\}$. However, half-lines are not endless.
Note that it can be shown that if an implicit representation $F(x)+G(y)=0$ should represent the union of curves $y=x$ and $y=x+1$, then in fact all the curves $y=x+a, a\in \mathbb Z$ are part of the solution. This can be seen from condition that if three of the points $(x_1,y_1),(x_1,y_2),(x_2,y_1),(x_2,y_2)$ satisfy $F(x)+G(y)=0$, then four of them do so. C.f. @DavidESpeyer comments in Closed curves of the form $F(x)+G(y)=0$
Following the comments of @JyrkiLahtonen, we can find for example the equation (I switched the role of $x$ and $y$ for the convenience of plotting the graph) $$ y+(1-y)^4 = cos(x), $$ which represents two disjoint endless curves as shown in the picture: