The point is that sometimes, different models (for the same data) can lead to likelihood functions which differ by a multiplicative constant, but the information content must clearly be the same. An example:
We model $n$ independent Bernoulli experiments, leading to data $X_1, \dots, X_n$, each with a Bernoulli distribution with (probability) parameter $p$. This leads to the likelihood function
$$
\prod_{i=1}^n p^{x_i} (1-p)^{1-x_i}
$$
Or we can summarize the data by the binomially distributed variable $Y=X_1+X_2+\dotsm+X_n$, which has a binomial distribution, leading to the likelihood function
$$
\binom{n}{y} p^y (1-p)^{n-y}
$$
which, as a function of the unknown parameter $p$, is proportional to the former likelihood function. The two likelihood functions clearly contains the same information, and should lead to the same inferences!
And indeed, by definition, they are considered the same likelihood function.
Another viewpoint: observe that when the likelihood functions are used in Bayes theorem, as needed for bayesian analysis, such multiplicative constants simply cancel! so they are clearly irrelevant to bayesian inference. Likewise, it will cancel when calculating likelihood ratios, as used in optimal hypothesis tests (Neyman-Pearson lemma.) And it will have no influence on the value of maximum likelihood estimators. So we can see that in much of frequentist inference it cannot play a role.
We can argue from still another viewpoint. The Bernoulli probability function (hereafter we use the term "density") above is really a density with respect to counting measure, that is, the measure on the non-negative integers with mass one for each non-negative integer. But we could have defined a density with respect to some other dominating measure. In this example this will seem (and is) artificial, but in larger spaces (function spaces) it is really fundamental! Let us, for the purpose of illustration, use the specific geometric distribution, written $\lambda$, with $\lambda(0)=1/2$, $\lambda(1)=1/4$, $\lambda(2)=1/8$ and so on. Then the density of the Bernoulli distribution with respect to $\lambda$ is given by
$$
f_{\lambda}(x) = p^x (1-p)^{1-x}\cdot 2^{x+1}
$$
meaning that $$
P(X=x)= f_\lambda(x) \cdot \lambda(x)
$$
With this new, dominating, measure, the likelihood function becomes (with notation from above)
$$
\prod_{i=1}^n p^{x_i} (1-p)^{1-x_i} 2^{x_i+1} = p^y (1-p)^{n-y} 2^{y+n}
$$
note the extra factor $2^{y+n}$. So when changing the dominating measure used in the definition of the likelihood function, there arises a new multiplicative constant, which does not depend on the unknown parameter $p$, and is clearly irrelevant. That is another way to see how multiplicative constants must be irrelevant. This argument can be generalized using Radon-Nikodym derivatives (as the argument above is an example of.)
