Why does a nowhere vanishing section of $\omega_{E/S}$ induce an isomorphism $\mathcal O_E \cong \Omega^1_{E/S}$?

I am reading Arithmetic Moduli of Elliptic Curves by Katz and Mazur, and I have a question about the beginning of Chapter 2, §2. Let $f:E\to S$ be an elliptic curve. Since the sheaf of relative diffrentials $\Omega^1_{E/S}$ is an invertible sheaf on $E$, one defines $$\omega_{E/S}=f_*\Omega^1_{E/S},$$ which is an invertible sheaf on $S$ (using Grothendieck–Serre duality). Now choose, Zariski locally on $S$, a basis $$\omega\in \Gamma(S,\omega_{E/S}).$$ By the definition of the pushforward, this is also a global section of $\Omega^1_{E/S}$ on $E$. The authors then state that this nowhere vanishing section determines an isomorphism $$\mathcal O_E \xrightarrow{\sim} \Omega^1_{E/S},$$ which i suspect is given by multiplication by $\omega$. My confusion is the following. A section of an invertible sheaf induces an isomorphism with the structure sheaf if and only if it is nowhere vanishing on the space where the sheaf lives, i.e., on $E$. However, I only know that $\omega$ is a local generator of the invertible sheaf $\omega_{E/S}$ on $S$. Why does this imply that the corresponding section of $\Omega^1_{E/S}$ is nowhere vanishing on $E$? Equivalently, how can one see explicitly that, for every point $x\in E$, the image of $\omega$ generates the stalk $(\Omega^1_{E/S})_x$?