As described, you use the weights $w^i = p(x^i) / g(x^i)$. We're dividing by something here: when can $g(x) = 0$? In order for $g(x) = 0$ we need to have sampled an $x_0$ such that $g(x_0) = 0$. But the density at $x_0$ is exactly $0$, so the probability (under $g$) of the set $\{x : g(x) = 0\}$ is exactly $0$. This means with probability $1$ we will never see $g = 0$ in these weights, so we don't have to worry about this.
What if instead we use some other weighting like the one you propose where we divide by $x$? Maybe that would cancel things out in the integral, but we may easily have $P_g(X = 0) > 0$, in which case the weights have a positive probability of being infinite/undefined, or even if $P_g(X = 0)$ the ratio with $x$ in the denominator may be very poorly behaved. For example, in what ought to be one of the friendliest of cases, when $X$ is Gaussian $1 / X$ does not have a finite mean.
Furthermore, we aren't even trying to cancel out the $x$. We instead want to change what we're effectively taking the expectation of in the first place, so having a $\frac 1x$ term wouldn't even help with that.
We have
$$
\frac 1m \sum_{i=1}^m x^{(i)} \to_p E_g(X) = \int_{\mathbb R} x g(x)\,\text dx.
$$
If we instead just use $w^i \cdot h(x^i)$ we have
$$
\frac 1m \sum_{i=1}^m \frac{p(x^i)}{g(x^i)} h(x^i) \to_p E_g\left[ \frac{p(X)}{g(X)} h(X)\right] = \int_{\mathbb R} p(x) h(x) \,\text dx = \mu.
$$
We never had any inconvenient $x^{(i)}$s that we needed to cancel out anyway.
