Are "pushforward" and "pullback by zero section" of diffrentials isomorphic in elliptic curves?

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Let S be a scheme, $p:E\rightarrow S$ be an elliptic curve with zero section $e:S\rightarrow E$. My question is: Are $p_*(\Omega_{E/S}^1)$ and $e^*(\Omega_{E/S}^1)$ isomorphic? I think it's right because I saw both definitions in different notes and they seem to represent the same thing.

Thank you!

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Yes, this is true.

Namely, note that [1, §4.2 Proposition 2] one has an isomorphism of $\mathcal{O}_E$-modules

$$p^\ast e^\ast\Omega^1_{E/S}\cong \Omega^1_{E/S}$$

Applying $p_\ast$ to both sides gives an isomorphism

$$p_\ast(p^\ast e^\ast \Omega^1_{E/S})\cong p_\ast \Omega^1_{E/S}$$

But, by the projection formula

$$\begin{aligned} p_\ast(p^\ast e^\ast\Omega^1_{E/S}) &=p_\ast(p^\ast e^\ast\Omega^1_{E/S}\otimes_{\mathcal{O}_E}\mathcal{O}_E)\\ &=e^\ast\Omega^1_{E/S}\otimes_{\mathcal{O}_S} p_\ast(\mathcal{O}_E)\\ &= e^\ast\Omega^1_{E/S}\otimes_{\mathcal{O}_S}\mathcal{O}_S\\ &= e^\ast \Omega^1_{E/S}\end{aligned}$$

where we have used the fact that since $p:E\to S$ is a smooth proper curve with geometrically connected fibers that $p_\ast\mathcal{O}_E\cong\mathcal{O}_S$.

[1] Bosch, S., Lütkebohmert, W. and Raynaud, M., 2012. Néron models (Vol. 21). Springer Science & Business Media.