I have come across the following identity for the delta function
$$\delta[g(x)] = \sum_i \frac{\delta(x-x_i)}{|\frac{d g}{dx}|_{x=x_i}} $$
$x_i$ are the roots of g. However, believe there is something odd about it. I don't understand how this preserves the units of the equation. For instance, suppose I want to do the following integration
$$\int f(\theta) \delta(E(\theta)) d\theta = \int f(\theta) \sum_i \frac{\delta(\theta-\theta_i)}{|\frac{d E}{d\theta}|_{\theta=\theta_i}} d\theta$$
where E is energy and $\theta$ in an angle, then the units of the right-hand side have an additional factor of (1/energy) with respect to the left-hand side
First, assume that $g:\mathbb R \to\mathbb R$ is a diffeomorphim. Then, for $\varphi$ a smooth compactly supported function, you have :
$$\begin{split} \int_{\mathbb R} \varphi(x) \delta(g(x)) d x &= \int_{\mathbb R}|\varphi(g^{-1}(y)) |(g^{-1})'(y) \delta(y) dy \\ &=\varphi(g^{-1}(0)) \times |(g^{-1})'(0)| \end{split}$$ And therefore:
$$\delta(g(x)) = \frac{1}{|g'(g^{-1}(0))|} \delta(x - g^{-1}(0))$$
In general, if $g$ is nice enough, you can cut the first integral into intervals on which $g$ is strictly monotonous, which gives your first formula.
As for the dimensional analysis, remember that the dimension of $\delta(x)$ is fixed by $\int \delta(x) dx = 1$. The missing $E^{-1}$ factor comes from the delta.