How is $\delta(at+b)=\frac{1}{|a|}\delta(t+b/a)$?
First of all it seems useful to establish what we mean by an equation like

$$\delta(at+b)=\frac{1}{|a|}\delta(t+b/a),\qquad a\neq 0\tag{1}$$

Since the Dirac impulse $\delta(t)$ is a distribution, Eq. $(1)$ only makes sense if interpreted as

$$\int_{-\infty}^{\infty}\delta(at+b)\phi(t)dt=\frac{1}{|a|}\int_{-\infty}^{\infty}\delta(t+b/a)\phi(t)dt\tag{2}$$

where $\phi(t)$ is a so-called test function, by which we generally mean that it has derivatives of any order, and that for $|t|\to\infty$ it tends to zero sufficiently rapidly.

In order to prove $(1)$, it is sufficient to show that

$$\delta(at)=\frac{1}{|a|}\delta(t),\qquad a\neq 0\tag{3}$$

because substituting $t+t_0$ for $t$ in $(3)$ results in

$$\delta(a(t+t_0))=\frac{1}{|a|}\delta(t+t_0),\qquad a\neq 0\tag{4}$$

which is equal to $(1)$ for $t_0=b/a$.

Proving $(3)$ means that we need to show that

$$\int_{-\infty}^{\infty}\delta(at)\phi(t)dt=\frac{1}{|a|}\int_{-\infty}^{\infty}\delta(t)\phi(t)dt,\qquad a\neq 0\tag{5}$$

Eq. $(5)$ can be shown in a straightforward manner by the variable substitution $at=\tau$, which results in $a\,dt=d\tau$:

$$\begin{align}\int_{-\infty}^{\infty}\delta(at)\phi(t)dt=\begin{cases}\displaystyle\frac{1}{a}\int_{-\infty}^{\infty}\delta(\tau)\phi(\tau/a)d\tau,&amp;a&gt;0\\\displaystyle\frac{1}{a}\int_{\infty}^{-\infty}\delta(\tau)\phi(\tau/a)d\tau,&amp;a&lt;0\end{cases}\tag{6}\end{align}$$

Since

$$\frac{1}{a}\int_{\infty}^{-\infty}\delta(\tau)\phi(\tau/a)d\tau=-\frac{1}{a}\int_{-\infty}^{\infty}\delta(\tau)\phi(\tau/a)d\tau\tag{7}$$

the result $(6)$ can be summarized as

$$\int_{-\infty}^{\infty}\delta(at)\phi(t)dt=\frac{1}{|a|}\int_{-\infty}^{\infty}\delta(t)\phi(t/a)dt,\qquad a\neq 0\tag{8}$$

which is equivalent to $(5)$ because

$$\begin{align}\frac{1}{|a|}\int_{-\infty}^{\infty}\delta(t)\phi(t/a)dt&amp;=\frac{1}{|a|}\int_{-\infty}^{\infty}\delta(t)\phi(t)dt\\&amp;=\frac{1}{|a|}\phi(0),\qquad a\neq 0\end{align}\tag{9}$$

Consequently, $(8)$ and $(9)$ prove $(3)$, which in turn proves $(4)$ and $(1)$.

Thanks to Laurent Duval for commenting and motivating me to rewrite this answer with a bit more (engineering) rigor.