When is the conditional mean of potential outcomes linear in the propensity score?

Consider an outcome $$Y$$, treatment $$D$$, and set of covariates $$X$$. The outcome is real-valued, the treatment has support $$\{0,1\}$$, and the set of covariates is a collection of binary variables with support $$\{0,1\}$$. Furthermore, let $$Y(d)$$ denote potential outcome of $$Y$$ if $$D=d$$ for $$d\in\{0,1\}$$, and note that $$Y=Y(1)D+Y(0)(1-D)$$. Now, assume the following.

Assumption 1 (Mean Independence). $$E(Y(d)\mid X,D)=E(Y(d)\mid X)$$ for $$d\in\{0,1\}$$.

In this context, how strong is the following assumption?

Assumption 2. $$E(Y(d)\mid X)=\alpha_d+\beta_d\cdot p(X)$$ for $$d\in\{0,1\}$$, where $$\{(\alpha_1,\beta_1),(\alpha_0,\beta_0)\}$$ are pairs of real numbers (i.e., scalars) and $$p(X)=P(D=1\mid X)$$ is the propensity score.

In other words, how strong is the assumption that the conditional means of the potential outcomes are linear in the propensity score? I am wondering if there are some easy-to-understand assumptions that imply Assumption 2. This is an important assumption that is used in the causal inference literature for adjustment on covariates (see, e.g., corollary 4.3 in [1]). Or is the assumption guaranteed already? For example, can it be derived from the simple fact that $$E(Y(d)\mid X)$$ and $$p(X)$$ are linear in $$X$$ because $$X$$ is a collection of binary variables (meaning that the model is saturated)?

1. Rosenbaum, P. R., & Rubin, D. B. (1983). The central role of the propensity score in observational studies for causal effects. Biometrika, 70(1), 41-55.

This is a fairly strong assumption and one not likely to be true in practice. That assumption being true justifies the use of a linear regression of $$Y$$ on $$D$$ and $$p(X)$$ for estimating the causal effect of $$D$$ on $$Y$$. In practice, because this is assumption is rarely true, this method is not a valid way to reduce the bias due to confounding by $$X$$. Matching, weighting, and stratification on the propensity score do not require this assumption and therefore are better used for estimating the effect of $$D$$. If using regression adjustment on the propensity score, one should use a flexible model of $$Y$$ on $$D$$ and $$p(X)$$, such as a spline model, which is more likely to capture the true relationship.
Let $$E[Y(0)|X] = E[Y(1)|X] = E[Y|X] = .2X_1 + .3X_2$$ (i.e., no treatment effect). Also let $$p(X)=.3X_1+.2X_2$$ There are no scalars $$\alpha$$ and $$\beta$$ such that $$E[Y|X] = .2X_1 + .3X_2 = \alpha + \beta(.3X_1+.2X_2) = \alpha + \beta \cdot p(X)$$So, the assumption fails.
The expectation of $$Y$$ given $$X$$ is a function of $$X.$$ Hence if $$X$$ takes value in $$\{0,1\}$$ then there always exists $$\alpha$$ and $$\beta$$ such that $$E[Y|X]=\alpha+\beta X$$ for the mere reason that you can alway draw a line between two given points. Hence if $$X$$ is univariate, both $$p(X)$$ and $$E[Y|X]$$ are linear functions fo $$X.$$
Now if $$X$$ is multivariate each components belonging to $$\{0,1\}$$ again things get a little more messy. Consider $$X$$ is a 2-dimensional vector. You can write $$E[Y|X]=E[Y|X_1=0,X_2=0]+ X_1E[Y|X_1=1,X_2=0]+ X_2E[Y|X_1=1,X_2=0]+ X_1X_2E[Y|X_1=1,X_2=1].$$ While the first three terms of the above RHS are linear function of $$X$$, the last one is not.
In the complete general case, your "linear" representation needs cross effects -which can be expressed as products of components of $$X$$- to hold true.