I really have no idea how to approach this problem. Thinking about it informally, $s$ ignores its argument and returns itself applied to itself. So maybe something like:
$$ s = \lambda x \,. ss $$
Then I could solve for $s$ somehow. But I'm not sure at all how to approach this really.
Do you know the $Y$ combinator? It is defined by $$Y = \lambda f. (\lambda x. f (x x)) (\lambda x. f (x x)).$$ You can check yourself that $Y$ has the property that $Y f = f (Y f)$ for any term $f$. Since $Y f$ is a fixed point of $f$, one can use $Y$ to solve recursive relations such as yours.
If you want that $s = \lambda x. ss$, then $s$ must be a fixed point of the term $$f = \lambda y. \lambda x. y y$$ because $f s = (\lambda y. \lambda x. yy) s = \lambda x. ss = s$. Therefore, you can define $$s = Y f = (\lambda f. (\lambda x. f (x x)) (\lambda x. f (x x)))( \lambda y. \lambda x. y y).$$
