Is Griffiths simply wrong here? (Electrostatic Boundary Conditions)

Two ways of seeing that it’s right:

  1. Consider the case of no charge. Then nothing interesting is happening at the sheet, so the fields should be equal: both sides of the equation are zero.
  2. Gaus’s law: the sum of the fields going away, which is the outward flux, is given by the charge. Since $E_{below}$ is defined as towards the charge, it enters that calculation with a minus sign.

He’s picked a sign convention where the field upward is positive everywhere. That means that $E_{below}$ is defined such that a positive value means "the E vector points up" and a negative value means "the E vector points down". Which way the E vector points is given by the physics:

  • if there's positive charge on the surface and no external field, the E field below it will point down, hence $E_{below}$ will have a negative value.
  • If there's a large upward going external field, then in that case the E field points upward everywhere, and $E_{below}$ will be positive.

To put it another way, the length of the $E_{below}$ arrow in the picture isn't showing you the absolute magnitude; that has to come from somewhere else and is written as "$E_{below}$", a number. And it's not even showing you the direction, because if $E_{below}$ is less than zero, the actual E vector is pointing the other way. That arrow there is just defining a direction, like $\hat{x}$, $\hat{y}$ and $\hat{z}$.

Griffiths in correct. The flux through the top portion of the box is not just $E_{\small{top}}^\top$ but actually $\vec E\cdot (A\vec n) $ where $\hat n$ is perpendicular to the surface and points out: in your diagram, $\hat n$ would be along $+{\hat z}$ for the top portion of the flux calculation.

For the bottom of the box the $\hat n$ vector points along $-\hat z$ so the flux through through those sides of the box work out to $$ (E_{\small{top}}^{\top} - E_{\small{bottom}}^\top)A \tag{1} $$ when $\vec E_{\small{bottom}}$ points along $+\hat z$: the minus sign in (1) comes from the $-\hat z$ for the direction of the surface element $d\vec A$ at the bottom of the box.

Thus, ignoring the thin sides of the box because the flux through those is arbitrarily small, you have $$ (E_{\small{top}}^{\top} - E_{\small{bottom}}^\top)A=\frac{1}{\epsilon_0} \sigma A $$ and the area cancels out.