From exercise 1.13 in Atiyah Macdonald

If $f\in K[x]$ is an irreducible polynomial then there is a field $K\subseteq E$ such that $f$ has a root in $E$. Just take $E=K[x]/(f)$, this is a field because the ideal $(f)\subseteq K[x]$ is maximal, and the element $x+(f)$ is a root. By induction it follows that if $f_1,...,f_r$ is a finite number of irreducible polynomials in $K[x]$ then there is an extension field $E$ where all these polynomials have a root.

So now, suppose we indeed have an equation of the form:

$1=\sum_{i=1}^r y_i f_i(x_{f_i}) \ \ \ \ (*)$

Where $y_i\in A$ and $f_i\in S$. Take an extension field $E$ of $K$ where all the polynomials $f_1, f_2,...,f_r$ (which are polynomials in one variable) have roots, say, $\alpha_1,\alpha_2,...,\alpha_r$. Now we can think of $(*)$ as an equation of polynomials over $E$. And now substitute $x_{f_1}=\alpha_1, x_{f_2}=\alpha_2,...,x_{f_r}=\alpha_r$ and $x_f=0$ for all other $f\in S$ into $(*)$. We obtain $1=0$, a contradiction.