$Y_\theta$ := $(h(a,b) - E[h]) \cdot z$. $\forall a,b$
$$E[ Y_\theta] = E[(h(a,b) - E[h]) \cdot z] ..(1) $$ where $z_i = y_i - g(a_i,b_i)$; $g(a,b) = E[y | (a,b)]$ and $E[z] = 0$
How to evaluate equation $1$, using the tower property?
Is it possible to compute first over $y_i$ conditioned on $(a,b)$ and then over $(a,b)$
$$E[Y_\theta] = E[(h(a,b) - E[h]) \cdot z]$$
where $z_i = y_i - g(a_i,b_i)$, $g(a,b) = E[y | (a,b)]$, and $E[z] = 0$.
My Try!
\begin{align*} E[Y_\theta] &= E[(h(a,b) - E[h]) \cdot z] \\ &= E[(h(a,b) - E[h]) \cdot (y - g(a,b))] \\ &= E[h(a,b) y - h(a,b) g(a,b) - E[h] y + E[h] g(a,b)] \\ &= E[h(a,b) y] - E[h(a,b) g(a,b)] - E[E[h] y] + E[h] E[y] \\ &= E[h(a,b) y] - E[h(a,b) g(a,b)] - E[h] E[y] + E[h] E[y] \\ &= E[h(a,b) y] - E[h(a,b) g(a,b)] \\ \end{align*}
where we have used the linearity of expectation in the first step, and the definitions of $z$ and $g(a,b)$ in the second step.
Next, we use the tower property of conditional expectation to expand the first term:
\begin{align*} E[h(a,b) y] &= E[E[h(a,b) y | (a,b)]] \\ &= E[h(a,b) E[y | (a,b)]] \\ &= E[h(a,b) g(a,b)] \end{align*}
where we have used the fact that $h(a,b)$ is independent of $y$ given $(a,b)$, and $g(a,b) = E[y | (a,b)]$.
Substituting this back into the previous equation, we get:
\begin{align*} E[Y_\theta] &= E[h(a,b) y] - E[h(a,b) g(a,b)] \\ &= E[h(a,b) g(a,b)] - E[h(a,b) g(a,b)] \\ &= 0 \end{align*}