Prove that $\frac{e^x-1}{x}$ is bijective

I am trying to prove that this function is bijective, but I don't know how to do it.

$$f:x \mapsto \frac{e^x-1}{x}$$

For that, I try to use the fact that this function has necessarily a reciprocal if it is bijective.
So, I think I need to prove that :
$$f(x)=y \Leftrightarrow x = f^{-1}(y)$$
If I understand correctly, this means that :
$$\frac{e^x-1}{x} = y \Leftrightarrow x = \frac{y}{e^y-1}$$

But my problem is that I don't know how to do this, and I don't know if there is a more efficient way to solve the problem.
Can anyone help me solve this problem?

$\newcommand{\R}{\mathbb{R}}$ Let us add some more structure. Consider the function $f: \R\setminus\{0\} \to \R_{+}\setminus \{1\}$ defined by $$ f(x) = \frac{e^x -1}{x} $$ and define its extension $g:\R \to \R_{+}$ where $g(0) = 1$. We can easily show that $g$ is continuous at $x = 0$ by L'Hopital's rule.

Now consider the derivative of $g$,

$$ g'(x) = \frac{e^x(x-1) +1}{x^2} $$

We have $$ \lim_{x \to 0} g'(x) = \frac{1}{2} $$

Hence from Hermis14, $g'(0) = 1/2$.

Furthurmore, $g'(x)$ is always positive and $g$ is surjective, which means $g$ is bijective.

Therefore, $f$ is bijective.

It's useful to notice that $$\frac{e^x-1}{x} = \int_0^1 e^{x t} d t$$ so the function $f$ and all its derivatives are $>0$, with $\lim_{x\to -\infty} \frac{e^x-1}{x} = 0$, $\lim_{x\to \infty} \frac{e^x-1}{x} = \infty$.

Obs: The reciprocal $g(x) = \frac{1}{f(x)} = \frac{x}{e^x-1}$ is the generating function of the Bernoulli numbers