Please find below a short argument in the case of schemes. The answer is positive for algebraic spaces too; in that case it can be proven using the characterization: $X$ has the resolution property $\Leftrightarrow$ $X = X'/GL_n$ with $X'$ quasi-affine $+$ products of quasi-affines are quasi-affine $+$ a bit more work.
Lemma. Let $S$ be a separated scheme. Let $X$ and $Y$ be quasi-compact and quasi-separated schemes over $S$. Then if $X$ and $Y$ have the resolution property, so does $X \times_S Y$.
Proof. Let $X = U_1 \cup \ldots \cup U_n$ and $Y = V_1 \cup \ldots \cup V_m$ be affine open coverings. Choose finite type quasi-coherent ideals $\mathcal{I}_i \subset \mathcal{O}_X$ such that $U_i = X \setminus V(\mathcal{I}_i)$. Choose finite type quasi-coherent ideals $\mathcal{J}_j \subset \mathcal{O}_Y$ such that $V_j = Y \setminus V(\mathcal{J}_j)$. Denote $\mathcal{K}_{ij} = (\text{pr}_1^{-1}\mathcal{I}_i)(\text{pr}_2^{-1}\mathcal{J}_j) \subset \mathcal{O}_{X \times_S Y}$ the finite type quasi-coherent ideal generated by the product of the pullback of $\mathcal{I}_i$ and the pullback of $\mathcal{J}_j$. Then the affine open $U_i \times_S V_j$ is equal to $X \times_S Y \setminus V(\mathcal{K}_{ij})$. As $X$ and $Y$ have the resloution property there exist surjections $\mathcal{E}_i \to \mathcal{I}_i$ and $\mathcal{F}_j \to \mathcal{J}_j$ with $\mathcal{E}_i$ finite locally free on $X$ and $\mathcal{F}_j$ finite locally free on $Y$. Then we see that there is a surjection $\text{pr}_1^*\mathcal{E}_i \otimes \text{pr}_2^*\mathcal{F}_j \to \mathcal{K}_{ij}$. Since $X \times_S Y = \bigcup_{i = 1, \ldots, n, j = 1, \ldots, m} U_i \times_S V_j$ we conclude $X \times_S Y$ has the resolution property by Lemma Tag 0F89.