Can being schematic be checked on an atlas? Let $X\to Y$ be a map between algebraic stacks, and $U\to Y$ a smooth atlas of $Y$. Suppose we know that $X\times_Y U$ is a scheme, can we show that $X\to Y$ is schematic?
 A: No. Consider the morphism $pt \to BG$ and the smooth atlas $pt \to BG$, for a smooth group scheme $G$. Then $pt \times_{BG} pt = G$ is a scheme.  However, $pt \to BG$ is in general only representable by algebraic spaces, and may not be schematic. See this answer for an example.
However, in your situation, one can always conclude that $X \to Y$  will be representable (by algebraic spaces, not necessarily by schemes). This follows from the fact that algebraic spaces satisfy étale descent, and even fppf descent, see Stacks 04SJ. Strictly speaking to apply the cited reference you need to reduce to the case where $Y$ is a scheme first.  In more details:
Let $T \to Y$ be a morphism with $T$ a scheme and let $S = T \times_Y X$. We need to show that $S$ is an algebraic space. By assumption there exists a smooth atlas $Y_0 \to Y$ such that the base change $X_0 = Y_0 \times_Y X \to X$ is a smooth atlas with $X_0$ also a scheme. Base change everything along $Y_0 \to Y$ so that you get a cartesian square $$\require{AMScd}\begin{CD}
U     @>>>  V\\
@VVV        @VVV\\
S     @>>>  T
\end{CD}$$
where $V = T \times_Y Y_0$, $U = S \times_X X_0$, and the vertical maps are smooth surjections. Also, $U$ and $V$ are algebraic spaces; that's because $X_0 \to X$ and $Y_0 \to Y$ are representable (automatic from the definition of algebraic stack). In particular, the upper map is representable. In other words, $S \to T$ is representable "smooth-locally on the target" (i.e. becomes representable after base change along a smooth atlas). Now the Stacks Project reference applies and implies that $S \to T$ itself is representable, hence $S$ is an algebraic space.
