The assumption implies that \[\begin{aligned} E[Y_{1i}|D_{i} & =1,X_{i}]=E[Y_{1i}|D_{i}=0,X_{i}]=E[Y_{1i}|X_{i}]\\ E[Y_{0i}|D_{i} & =1,X_{i}]=E[Y_{0i}|D_{i}=0,X_{i}]=E[Y_{0i}|X_{i}]\end{aligned}\]
The \(ATT\) for \(X_{i}=x\) is given by \[\begin{aligned} E[Y_{1i}-Y_{0i}|D_{i}=1,X_{i}] & =E[Y_{1i}|D_{i}=1,X_{i}]-E[Y_{0i}|D_{i}=1,X_{i}]\\ & =E[Y_{i}|D_{i}=1,X_{i}]-E[Y_{0i}|D_{i}=0,X_{i}]\\ & =\underbrace{E[Y_{i}|D_{i}=1,X_{i}]}_{avg\ with\ X_{i}\ in\ treatment}-\underbrace{E[Y_{i}|D_{i}=0,X_{i}]}_{avg\ with\ X_{i}\ in\ control}\end{aligned}\]
The components in the last line are identified (can be estimated).
Intuition: Comparing the outcome across control and treatment groups after conditioning on \(X_{i}\)
ATT is given by \[\begin{aligned} ATT & =E[Y_{1i}-Y_{0i}|D_{i}=1]\\ & =\int E[Y_{1i}-Y_{0i}|D_{i}=1,X_{i}=x]f_{X_{i}}(x|D_{i}=1)dx\\ & =E[Y_{i}|D_{i}=1]-\int\left(E[Y_{i}|D_{i}=0,X_{i}=x]\right)f_{X_{i}}(x|D_{i}=1)\end{aligned}\]
ATE is \[\begin{aligned} ATE= & E[Y_{1i}-Y_{0i}]\\ = & \int E[Y_{1i}-Y_{0i}|X_{i}=x]f_{X_{i}}(x)dx\\ = & \int E[Y_{i}|D_{i}=1,X_{i}=x]f_{X_{i}}(x)dx\\ = & +\int E[Y_{i}|D_{i}=0,X_{i}=x]f_{X_{i}}(x)dx\end{aligned}\]