For any $a,b,\theta \in \mathbb{R}$ and $c > 0$, it holds $ F(\theta ) := \int_{ - \infty }^\infty {\Phi (\theta - a - bx)\phi (cx - \theta )\,{\rm d}x} = \frac{1}{c}\Phi \bigg(\frac{{(c - b)\theta - ac}}{{\sqrt {c^2 + b^2 } }}\bigg). $ (I confirmed the result numerically.)
Proof. First note that $ \frac{1}{c}\Phi \bigg(\frac{{(c - b)\theta - ac}}{{\sqrt {c^2 + b^2 } }}\bigg) = \frac{1}{c}{\rm P}\bigg[Z \le \frac{{(c - b)\theta - ac}}{{\sqrt {c^2 + b^2 } }}\bigg] = \frac{1}{c}{\rm P}\big[cX + bY \le (c - b)\theta - ac\big], $ where $X$, $Y$, and $Z$ are independent ${\rm N}(0,1)$ random variables (note that $\sqrt {c^2 + b^2 } Z$ and $cX+bY$ are identically distributed). By the law of total probability, conditioning on $Y$, we thus get $ \frac{1}{c}\Phi \bigg(\frac{{(c - b)\theta - ac}}{{\sqrt {c^2 + b^2 } }}\bigg) = \frac{1}{c}\int_{ - \infty }^\infty {{\rm P}\big[cX + by \le (c - b)\theta - ac\big]\phi (y)\,{\rm d}y}. $ A change of variable $y=cx-\theta$ then gives $ \frac{1}{c}\Phi \bigg(\frac{{(c - b)\theta - ac}}{{\sqrt {c^2 + b^2 } }}\bigg) = \frac{1}{c}\int_{ - \infty }^\infty {{\rm P}\big[cX + b(cx - \theta ) \le (c - b)\theta - ac\big]\phi (cx - \theta )c\,{\rm d}x} . $ A little algebra shows that the expression on the right is equal to $ \int_{ - \infty }^\infty {{\rm P}\big[X \le \theta - a -bx \big]\phi (cx - \theta )\,{\rm d}x}, $ and hence $ \frac{1}{c}\Phi \bigg(\frac{{(c - b)\theta - ac}}{{\sqrt {c^2 + b^2 } }}\bigg) = \int_{ - \infty }^\infty {\Phi (\theta - a - bx)\phi (cx - \theta )\,{\rm d}x}. $