It reduces the analysis to a (generally simpler) "standard" likelihood in which the location and scale parameters have been set to convenient values. There is a pitfall for the unwary, though: to obtain the general likelihood from the standard likelihood, you must introduce a factor that depends on the scale parameter.
A family $\mathcal F$ of probability distributions is a "location" family if, whenever the cumulative distribution function $F$ is in the family, then so is the function $x \to F(x-\mu).$ It is a "scale" family when $F\in \mathcal F$ implies the function $x\to F(x/\sigma)$ is in the family for any positive number $\sigma.$ (It is a location-scale family when both properties hold.)
We may take $\mu$ and $\sigma$ to be parameters of the family. Additional parameters, which I will just generically call $\theta,$ may be needed to specify members of the family uniquely.
In this setting, for each value of $\theta$ we may choose a "standard" distribution $F_\theta$ once and for all to represent it, making the general distribution in the family of the form
$$F_\theta(x;\mu,\sigma) = F_\theta\left(\frac{x-\mu}{\sigma}\right).$$
(This first shifts the location by $\mu$ and afterwards scales it by $\sigma.$ If we were to reverse the order--which is perfectly reasonable--the argument would instead be $x/\sigma - \mu.$)
Almost always, such families consist of absolutely continuous distributions. Let $f_\theta$ be the density associated with $F_\theta.$ Consequently, when $F$ is the distribution with parameters $(\mu,\sigma,\theta)$ its associated probability element is
$$\mathrm{d}F_\theta(x;\mu,\sigma) = \mathrm{d}F_\theta\left(\frac{x-\mu}{\sigma}\right) = \frac{1}{\sigma}f_\theta\left(\frac{x-\mu}{\sigma}\right)\,\mathrm{d}x.$$
Notice the appearance of the factor $1/\sigma.$ This subtlety is the reason for going through this analysis!
When a sample of (independent) data $\mathbf x = (x_1, \ldots, x_n)$ is observed from this family, its likelihood $\mathcal L$ is, by definition, its probability density. Since independence means probability densities multiply, we obtain
$$\mathcal{L}(\mu, \sigma, \theta; \mathbf x) = \prod_{i=1}^n \frac{1}{\sigma}f_\theta\left(\frac{x_i-\mu}{\sigma}\right)= \sigma^{-n}\prod_{i=1}^n f_\theta\left(\frac{x_i-\mu}{\sigma}\right) = \sigma^{-n} \mathcal{L}\left(0, 1, \theta; \frac{\mathbf x - \mu}{\sigma}\right) .$$
On the right, $\mathbf x - \mu$ is shorthand for $(x_1-\mu, \ldots, x_n-\mu).$
Here, then, is the result: suppose you have an expression for the likelihood for a standardized distribution where $\mu=0$ and $\sigma=1;$ say, some function $\Lambda(\theta;\mathbf x).$ (This plays the role of $\mathcal L$ on the right hand side just above.) Then the likelihood in full generality is
$$\mathcal{L}(\mu, \sigma, \theta; \mathbf x) = \sigma^{-n} \Lambda\left(\theta;\frac{\mathbf x - \mu}{\sigma}\right) .$$
That is, to obtain the general likelihood from the standard likelihood, (1) replace all $x_i$ by $(x_i-\mu)/\sigma$ and (2) multiply it by $\sigma^{-n}.$ Don't forget the second step!
Of course, in location-only families you can ignore $\sigma$ (set it to $1$) and in scale-only families you can ignore $\mu$ (set it to $0$).
Example
The Normal distribution family is a scale-location family (with no additional parameters). For the last century, most people have chosen the function
$$f(x) = \exp\left(-\frac{x^2}{2}\right)$$
(defined for all real numbers) for the standard density. Accordingly, using the rules of exponentiation, the likelihood of a dataset from the standard Normal distribution would be
$$\Lambda(\mathbf x) = \exp\left(-\sum_{i=1}^n \frac{x_i^2}{2}\right).$$
Therefore, the likelihood generally must be
$$\mathcal L(\mu,\sigma;\mathbf x) = \sigma^{-n} \exp\left(-\sum_{i=1}^n \frac{(x_i-\mu)^2}{2\sigma^2}\right).$$