We want to show the following:
$$ \mathbb{E}[Y \vert X, D=1] \overset{\sigma(x)-a.s.}{=} \mathbb{E}[\tilde{Y}(1) \vert X] $$
We can express the left-hand side as follows:
$$ \begin{align*}\mathbb{E}[Y \vert X, D=1]&=\int Y d\mu_{D=1, \sigma(X)} && \text{by Definition} \\&= \int \tilde{Y}(1)d\mu_{D=1, \sigma(X)} && \text{by SUTVA}\\ &= \int \tilde{Y}(1)d\mu_{\sigma(X)} && \text{by CIA} \\ &=\mathbb{E}[\tilde{Y}(1) \vert X]&& \text{by Definition}\end{align*} $$
By the conditional independence assumption:
$$ \begin{align*}\forall A \in \mathcal{F}, \quad \mu_{D=1, \sigma(X)}(\cdot, A)&= \frac{\mu_{\sigma(X)}(\cdot, A \cap D=1)}{\mu_{\sigma(X)}(\cdot, D=1)} && \text{by Definition}\\ &= \frac{\mu_{\sigma(X)}(\cdot, A)\mu_{\sigma(X)}(\cdot, D=1)}{\mu_{\sigma(X)}(\cdot, D=1)} && \text{by Independence}\\ &= \mu_{\sigma(X)}&& \text{by Definition}\end{align*} $$