Must all Lebesgue integrable functions really be invertible?

$f^{-1}(A)$ is the preimage of the set $A$, it exists even if $f$ is not invertible. For $f: X\to Y$, it is defined as

$$f^{-1}(A) = \{x\in X: f(x)\in A\}.$$

The notation is slightly confusing, I would say, but one gets used to it. It isn't really that bad of a notation, since, if $f$ is actually invertible and its inverse is $g$, then $g(A) = f^{-1}(A)$ (so the preimage of $A$ according to $f$ is the image of $A$ according to $f^{-1}$).

$f^{-1}$ here does not mean the inverse of $f$. It is a horrifically bad overuse of notation and can be very misleading. $f^{-1}(A)$ where $A$ is a set is the pre-image, i.e.

$$f^{-1}(A) = \{x\in X: f(x) \in A\}.$$

The pre-image always exists. If $f$ were invertible, then this would coincide with what you think; however the pre-image takes care of the multi-valued aspect.

Take for instance $f:[-1,2]\to\mathbb{R}$ defined by $f(x) = x^2$. This is not invertible as you can see by, e.g., $f(-1) = 1 = f(1)$. However

$$f^{-1}([0,1]) = \{x\in [-1,2]:f(x) \in[0,1]\}.$$

The set of $x$ that solve this are those $x$ in $[-1,1]$ since when you square them you get something in $[0,1]$.

This function is not invertible but it is continuous on a set of finite Lebesgue measure, hence Lebesgue integrable.

For any function $f : \mathbb R \rightarrow \mathbb R$, the set $f^{-1}( a, \infty)$ always exists, and it is defined by $$ \{ x: \in \mathbb R : f(x) > a \}, $$ although it is not always measurable.