Let $K$ be a field and $A$ a subring of $K$ with the following property: for every finite family $(\xi_i)_{1\leq i\leq n}$ of elements of $K$ there exists $\gamma\in A-\{0\}$ such that $\gamma\xi_i\in A$ for $1\leq i\leq n$ (a hypothesis which is always satisfied when $A$ is abelian and $K$ is the field of fractions of $A$).
First of all, if $A$ is only nonzero and abelian, then it is not necessarily an integral domain; therefore, how can it possess a field of fractions $K$? Surely they mean that $A$ is an integral domain and not only an abelian and nonzero ring, correct?
Now, if $K$ is the field of fractions of $A$, then $A$ can be identified a with its image in $K$: i.e. $A$ is a nonzero subring of the abelian field $K$. Therefore there exist $\gamma\in A$ such that $\gamma\ne0$. But why does it follow that $\gamma\xi_i\in A$ for all $1\leq i\leq n$?
Notational Reminder: "field" here means "skew field", i.e., multiplication need not be abelian. Also, "ring" means "ring with $1$" (or at least "ring with more than one element").
Claim. Let $K$ be a field, $A$ a an abelian subring and assume that there is no field $L$ with $A\subseteq L\subsetneq K$. Let $\xi_1,\ldots,\xi_n\in K$. Then there exists $\gamma\in A\setminus\{0\}$ with $\gamma \xi_i\in A$ for $1\le i\le n$.
Proof. Let $$L=\{\,\xi\in K\mid\exists \gamma\in A\setminus\{0\}\colon \gamma\xi\in A\,\}.$$ Clearly, $A\subseteq L$ (here we use $A\setminus\{0\}\ne\emptyset$). Assume $\xi_1,\xi_2\in L$. Let $\gamma_1,\gamma_2\in A\setminus\{0\}$ with $\gamma_1\xi_1,\gamma_1\xi_2\in A$. Then $\gamma:=\gamma_1\gamma_2\in A\setminus\{0\}$ and $$ \gamma(\xi_1-\xi_2)=\gamma_2\gamma_1\xi_1-\gamma_1\gamma_2\xi_2\in A$$ so that $L$ is a subgroup of $K$ under addition. Also, $$\gamma\xi_1\xi_2=\gamma_2\gamma_1\xi_1\xi_2=\gamma_1\xi_1\gamma_2\xi_2\in A$$ so that $L$ is closed under multiplication. Finally, if $\xi_1\ne 0$, then $\gamma_1\xi_1\in A\setminus\{0\}$ and $$ \gamma_1\xi\cdot\xi_1^{-1}=\gamma_1\in A.$$ It follows that $L$ is a subfield of $K$ and contains $A$. By assumption $L=K$. This solves the claim for $n=1$. Now let $\xi_1,\ldots,\xi_n\in K$. As just seen, there exist $\gamma_1,\ldots,\gamma_n\in A\setminus\{0\}$ with $\gamma_i\xi_i\in A$ for $1\le i\le n$. Then $\gamma:=\gamma_1\cdots\gamma_n\in A\setminus\{0\}$ and $$ \gamma\xi_i=\gamma_1\cdots\gamma_{i-1}\gamma_{i+1}\cdots\gamma_n\cdot\gamma_i\xi_i\in A,$$ as was to be shown. $\square$