Since $H$ is an everywhere differentiable function, we have
$$
\int x \, dH(x) = \int x H'(x)\,dx = \int x \Phi'\left(\frac{Ax+b}{\sqrt{2}}\right) \frac{A}{\sqrt{2}}\, dx.
$$
Let $u=\dfrac{Ax+b}{\sqrt{2}}$, so that $du = \dfrac{A\,dx}{\sqrt{2}}$ and $x= \dfrac{\sqrt{2}\; u - b}{A}$. Then the integral becomes
$$
\int \frac{\sqrt{2}\;u-b}{A} \Phi'(u) \, du.
$$
Then make this into
$$
\frac{\sqrt{2}}{A} \int u \Phi'(u)\,du - \frac{b}{A} \int \Phi'(u)\,du.
$$
The second integral above becomes $\Phi(u)+\text{constant}$. The first is
$$
\int u \Phi'(u)\,du = \int u e^{-u^2/2} \, du = \int e^{-w}\,dw = -e^{-w}+\text{constant}.
$$
Then convert back to an expression in $u$, then back to an expression in $x$.
If you had in mind a definite integral from $-\infty$ to $\infty$, then
$$
\int_{-\infty}^\infty \Phi'(u)\,du = 1
$$
and
$$
\int_{-\infty}^\infty u \Phi'(u) \, du = 0.
$$
The second integral is $0$ because you're integrating an odd function over an interval that is symmetric about $0$. The first is $1$ since you're integrating a probability density function.