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1 [made Community Wiki]

Here is an attempt to answer my own question, using the "division polynomials" of Kevin's and Jared's answers. It is probably the maximally naive idea, and I do not claim it works, though it's not clear to me that it can't. I've community wiki-ed this answer, as it's probably junk anyway ...

Fix $A,B\in \mathbb{Z}$, obtaining an elliptic curve $E$ over $S=Spec \mathbb{Z}[\Delta^{-1}]$, and a prime $p\geq 5$ not dividing the discriminant $\Delta=\Delta(A,B)$. The division polynomial $\psi_p(t)$ is supposed to be a polynomial of degree $d=(p^2-1)/2$ over $\mathbb{Z}$, whose roots are the values $t(P)=x(P)/z(P)$ as $P$ ranges over the points of exact order $p$ in $E$.

Turn $\psi_p$ into a homogeneous polynomial $g$ of degree $d$ in $\mathbb{Z}[x,y,z]$, so that $g(x,y,1)=\psi_p(x)$. The polynomial deterimes a curve $C=(g)$ in $P^2/S$, and thus a closed subscheme $D=E\cap C$ of $E$. Over $\mathbb{Z}[\Delta^{-1},p^{-1}]$, $D$ should consist of the points of exact order $p$ on $E$ (with multiplicity $1$), together with the basepoint of $E$ with multiplicity $d$.

Claim. $D$ is an effective Cartier divisor on $E/S$, of degree $3d$.

Proof. I don't know. I need to prove that $D\to S$ is flat, the main concern being the behavior over $\mathbb{Z}_{(p)}$. I don't even know if this is really plausible in general.

Let's pretend we somehow know $D$ is an effective Cartier divisor on $E$ relative to the base $S$. There is another relative Cartier divisor, namely $$D' = E[p] \quad + \quad (d-1)[0]$$ where $E[p]$ is the $p$-torsion subgroup scheme of $E$, and $[0]$ is the degree one divisor of the basepoint of $E$. It seems clear that away from characteristic $p$, the divisors $D$ and $D'$ are equal. Equality of effective divisors on a smooth curve is a closed condition, so they should be equal over all of $S$.

The divisor I want is thus $D''=D-d[0]=D'-d[0]$ (which is still effective). Then it's really easy to find a homogeneous polynomial $h$ of degree $2d/3$ which defines $D''$; because of the form of the Weierstrass equation, you can produce it from $g$ by hand, and if my claim is true you can produce it globally, i.e., with coefficients in $\mathbb{Z}[A,B]$.

I've carried out this out in the case $p=5$, and it appears to "work". That is, I get an answer which appears sane for general $A$ and $B$, and which for some explicit cases I've tried of $A,B\in \mathbb{Z}$ appears to give me a flat $D\to S$. For instance, if $E/\mathbb{Z}[6^{-1}]$ is the curve with $(A,B)=(0,1)$ (which reduces to a supersingular curve at $p=5$), I find $$h= 729z^{8}-1350x^4z^4+360x^6z^2+5x^8.$$