First let's clarify notations. $f(x)=x^e \pmod N$ is non-standard, hesitating between
- $f(x)\equiv x^e\pmod N$ , meaning $N$ divides $x^e-f(x)$
- $f(x)=x^e\bmod N$ , additionally specifying that $f(x)$ is a particular member of a finite set of $N$ elements, equivalently integers in range $[0,N)$ or $\mathbb Z_N$ .
What's meant is $f(x)=x^e\bmod N$.
Any injective function from a finite set to that same set is a bijection, thus we only need to prove that for any integers $x$ and $y$ in $[0,N)$, if $f(x)=f(y)$ then $x=y$. We do that in the following.
Assume $f(x)=f(y)$. By definition of $f$ that means $(x^e\bmod N)=(y^e\bmod N)$. That implies $N$ divides $x^e-y^e$. That implies any prime factor $p$ of $N$ divides $x^e-y^e$, that is $x^e\equiv y^e\pmod p$.
It is hypothesized $1=\gcd(e,(p-1)\cdot(q-1))$. Therefore $e$ and $p-1$ are coprime, the multiplicative inverse $d_p$ of $e$ in $\mathbb Z_{p-1}$ is well defined, and there exists a positive integer $d_p$ and a non-negative integer $k$ such that $e\cdot d_p=1+k\cdot(p-1)$.
Raising $x^e\equiv y^e\pmod p$ to that power $d_p$, we get that $(x^e)^{d_p}\equiv (y^e)^{d_p}\pmod p$; thus $x^{e\cdot d_p}\equiv y^{e\cdot d_p}\pmod p$; thus $x^{1+k\cdot(p-1)}\equiv y^{1+k\cdot(p-1)}\pmod p$.
For any prime $p$ and any integer $x$, Fermat's little theorem states that $x^p\equiv x\pmod p$. That allows to prove by induction on $k$ that for any non-negative integer $k$, $x^{1+k\cdot(p-1)}\equiv x\pmod p$.
Thus $x^{1+k\cdot(p-1)}\equiv y^{1+k\cdot(p-1)}\pmod p$ becomes $x\equiv y\pmod p$, that is $p$ divides $x-y$. That holds for any prime $p$ that divides $N$.
Hypothesizing that $N=p\cdot q$ with $p$ and $q$ distinct primes, we have shown that $f(x)=f(y)$ implies that distinct primes $p$ and $q$ divide $x-y$. It follows that $p\cdot q$ divides $x-y$, that is $x\equiv y\pmod N$, that is $x=y$ given that both belong to the set $[0,N)$; that completes our proof.