$\textbf{Moral:}$ Trying to compute the discriminant of this ring.
I claim that a basis for this ring is $\{1, \sqrt{d}, \sqrt{e}, \sqrt{ed}\}$ but I've been told by a peer that this may not be the case. However, I don't see anything wrong with my proof below:
First, note that $\{1,\sqrt{d}\}$ is basis for $\mathbb Z[\sqrt{d}][\sqrt{e}]$ over $\mathbb Z[\sqrt{e}]$: $\sqrt{d}$ is integral over $\mathbb Z[\sqrt{e}]$ since it is a root of the monic polynomial $m(x) = x^2 - d \in \mathbb Z[\sqrt{e}][x]$ and hence $\mathbb Z[\sqrt{d}][\sqrt{e}]$ is finitely generated over $\mathbb Z[\sqrt{e}]$ with basis $\{1,\sqrt{d}\}$ $(\dagger)$ (linear independence follows since if $e$ and $d$ are squarefree, so is $ed$). Similarly, $\sqrt{e}$ is integral over $\mathbb Z$ since it is a root of $m(x) = x^2 - e \in \mathbb Z[x]$. Hence $\mathbb Z[\sqrt{e}]$ is a finitely generated $\mathbb Z$-module with basis $\{1,\sqrt{e}\}$ $(\ast)$. Now, let $c_1,\dots,c_4 \in \mathbb Z$ and suppose that $c_1 + c_2\sqrt{e} + c_3\sqrt{d} + c_4\sqrt{e}\sqrt{d} = 0$. If we consider this linear combination as a linear combination of the basis $\{1,\sqrt{d}\}$ for $\mathbb Z[\sqrt{d}][\sqrt{e}]$ over $\mathbb Z[\sqrt{e}]$ then we have that \begin{equation} 1\cdot \left[ c_1 + c_2\sqrt{e} \right] + \left[ c_3 + c_4\sqrt{e} \right]\sqrt{d} = 0 \label{eq:1} \end{equation} By $(\dagger)$ the equation above implies that $c_1 + c_2\sqrt{e} = 0$ and that $c_3 + c_4\sqrt{e} = 0$. But then by $(\ast)$ we have that $c_1 = \dots = c_4 =0 $. It is clear that $\operatorname{Span} \mathcal A \subseteq \mathbb Z[\sqrt{d},\sqrt{e}]$. Now let $\alpha \in \mathbb Z[\sqrt{e},\sqrt{d}]$. By $(\dagger)$ we have that $\alpha = \beta_1 + \beta_2\sqrt{d}$ for some $\beta_1,\beta_2 \in \mathbb Z[\sqrt{e}]$. By $(\ast)$ we have that $\beta_1 = r_1 + r_2\sqrt{e}$ and $\beta_2 = r_3 + r_4\sqrt{e}$ for some $r_i \in \mathbb Z$. Substituting $\beta_1$ and $\beta_2$ into our equation for $\alpha$ implies that $\alpha \in \operatorname{Span} \mathcal{A}$, as desired. QED
