One possible approach to this is to divide the problem into two and figure out what conditions are needed for the two parts.

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First assume that $\pi_*E$ is locally free of the same rank as $E$ (say $r$). By adjointness of $\pi^*$ and $\pi_*$ we have a natural map
$$
\nu:\pi^*\pi_*E\to E
$$
**If** $E|_F$ is trivial, then $\nu$ is surjective (this can be checked locally, to do that at $x\in X$, take a set of local generators of $\pi_*E$ near $\pi(x)$. Their pre-images in $\pi^*\pi_*E$ will map to a set of generators in $E_x$). 
But then $\nu$ actually has to be an isomorphism, because $\pi^*\pi_*E$ and $E$ has the same rank, so it is an isomorphism at the generic points of $X$ and hence the kernel of $\nu$ is a torsion sheaf. Since $\pi^*\pi_*E$ is torsion-free this shows that $\nu$ is injective and hence an isomorphism.

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OK, so how can we guarantee that $\pi_*E$ is locally free of the same rank as $E$?

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We certainly need that $E|_F$ is trivial, but probably a little more. 

Let me say, without exploring whether weaker conditions suffice, that the following works:

**If** $Y$ is integral and all fibers of $\pi$ are reduced, then this is OK. This is essentially Grauert's theorem (see Hartshorne, III.12.9). 

(The condition on the fibers is to ensure that $h^0(F, E|_F)$ is constant. See Mohan's comment to Qing Liu's answer). 

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So we have the following:

> Let $\pi:X\to Y$ be a flat projective morphism and assume that $Y$ is integral and all fibers of $\pi$ are reduced. 
Further let $E$ be a locally free sheaf on $X$ such that $E$ restricted to any fiber of $\pi$ is trivial. Then $$\pi^*\pi_*E\simeq E.$$

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**Note** some of the assumptions you made or was willing to make are not necessary. 
You don't need $Y$ to be smooth or the higher cohomology of $\mathscr O_F$ to vanish.