Is being an algebraic space an Etale local property

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I was wondering the following. Let $S$ be a scheme, and let $\mathcal{F}$ be a sheaf on the big Etale site $(Sch/S)_{Et}$. Suppose there exists an Etale covering $(U_i)_i$ of $S$ such that the inverse image sheaf $\mathcal{F}|_{U_i}$ on $(Sch/U_i)_{Et}$ is an algebraic space over $U_i$. Is $\mathcal{F}$ then itself an algebraic space on $S$?

I know it is true if the covering is a Zariski covering. And it must be true considering the intuitive definition of algebraic spaces. But from the definitions on Stacks project, as well as the constructions out of Etale equivalence relations, it is not clear at all.

As an application, consider Etale vector bundles. Those are quasi-coherent $\mathcal{O}_S$-modules $\mathcal{F}$ (on the big site) such that there exists an Etale covering $(U_i)_i$ of $S$ such that $\mathcal{F}|_{U_i}=\mathcal{O}_{U_i}^{n_i}$ for some $n_i$ (i.e. they are locally free). Those are locally algebraic spaces: the restriction to $U_i$ is representable by the scheme $U_i\times Spec(\mathbb{Z}[x_1,...,x_{n_i}])$. So by above statement, it follows that the sheaf is an algebraic space.