Assume that $Z = X + Y$ where $X$ and $Y$ are two independent random variables with continuous pdfs. Is there, in general, a simple solution for the following:
$ \mathbb E_Y[\mathbb P(Y>0|Z)] $
The expection $\mathbb E_Y$ is taken over the $Y$ variable. To be more precise, let's assume the function $R(z)=\mathbb P(Y>0|Z=z)$ is well defined on $\mathbb R$. We want to find $ Q(x) = \mathbb E[R(x+Y)]$ for any $x\in \mathbb R$.
As a special case, when $ X \sim \mathcal N(0,\sigma_x^2)$ and $ Y \sim \mathcal N(0,\sigma_y^2)$ then $\mathbb P(Y>0|z)= \frac{1}{2}$ thus $ \mathbb E_y[\mathbb P(y>0|z)] = \frac{1}{2}$.
Maybe I did not understood your second statement but is this the result you are looking for? $$Q(x)= \mathbb{E}(R(x+Y))= \mathbb{E}(\mathbb{P}(Y>0 \vert X+Y=x+Y))= \mathbb{E}(\mathbb{P}(Y>0))$$