The key identity here is
$ S(x)=(x^p-x)^{p-1}=x^{p-1}(x^{p-1}-1) $
so that the polynomial $S$ above can be written both as a function if $x^p-x$ and a function of $x^{p-1}$. We show below that the fixed field is ${\mathbb F}_p(S(x))$.
Let $f(x)$ be a rational fraction fixed by $K$. We can write $f(x)=c\frac{u(x)}{v(x)}$ where $c$ is a constant and $u$ and $v$ are coprime unitary polynomials in $x$. By hypothesis, $f$ is fixed by $x \mapsto dx+a$, so $f(dx+a)=f(x)$ and hence $u(dx+a)v(x)=v(dx+a)u(x)$. By Gauss' lemma, $u(x)$ divides $u(dx+a)$.
Let $m$ be the degree of $u$. Then $d^mu(x)$ divides $u(dx+a)$. But since those two polynomials share the same degree and leading term (equal to 1), they must be equal. So $u(dx+a)=d^mu(x)$ for any $d,a$.
Taking $a=0$ and $d$ primitive as in Hurkyl's answer, we see that there is a polynomial $G$ such that $u(x)=G(x^{p-1})$.
Similarly, taking $a=d=1$, we see that $u(x+1)=u(x)$. We deduce that $w=u(x)-u(0)$ is identically zero on ${\mathbb F}_p$, so $w$ is a multiple of
$ \prod_{q\in {\mathbb F}_p}(x-q)=x^{p}-x $ By induction, it is easy to see then that there is a polynomial $H$ such that $u(x)=H(x^p-x)$.
Therefore $ u(x)=G(x^{p-1})=H(x^p-x) $
So ${\sf deg}(u)=(p-1){\sf deg}(G)=p{\sf deg}(H)$. So ${\sf deg}(u)$ is a multiple of $p(p-1)$. If $u$ is nonconstant, we can make an euclidian division of $u$ by $S$, and by induction we have $u(x)\in {\mathbb F}_p(S(x))$. Similarly $v(x)\in {\mathbb F}_p(S(x))$, so $f(x) \in {\mathbb F}_p(S(x))$, qed.