Inverse Continuous Wavelet Transform derivation?
Summary: the dual wavelet's role is analogous to that of $e^{j\omega t}$; it undoes the wavelet's convolving with the signal (integrated inner product). The main intricacy's indeed in normalization; there's no exact answer without digging in math (done below) - but intuitively, the seemingly matter-worsening $a^{-2}$ is due partly to the dilating nature of the wavelet, i.e. $\psi(t/a)$, which upon inversion requires downscaling not only to properly preserve energy, but reconstruct correctly by undoing the inner product.
Below are more &quot;proofs&quot; rather than &quot;derivations&quot;, as they work backwards to show equality with $f(t)$, but I term them latter as full-fledged proofs are beyond the scope of this answer.

Derivation 1: Convolution Theorem
Pasted for convenience the CWT and iCWT
$$
W_f(a, b) =   \int_{-\infty}^{\infty} f(t) \frac{1}{|a|^{1/2}} \psi \left(\frac{t - b}{a}\right) dt \tag{0}
$$
$$ 
f(t) = \frac{1}{C_{\psi}} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} W_f(a,b) \frac{1}{|a|^{1/2}} \psi \left(\frac{t - b}{a}\right) db \frac{da}{a^2}, \tag{1}
$$
Begin by rewriting the CWT as a convolution: $W_f(a, b) = f \star {\bar\psi_a}(b)$, where $\psi_a(t)=a^{-1/2}\psi(t/a)$. Inserting into $(1)$, and letting the right side equal $b(t)$,
$$
\begin{align}
b(t) &amp;= \frac{1}{C_\psi} \int_0^{\infty} W_f(.,a) \star \psi_a(t) \frac{da}{a^2} \\
     &amp;= \frac{1}{C_\psi} \int_0^{\infty} f \star {\bar\psi_a}(b) \star \psi_a(t) \frac{da}{a^2} \tag{2}
\end{align}
$$
where we also expressed the $db$ integral as a convolution, and $"."$ indicates the variable over which convolution is computed. We prove $b=f$ by showing their Fourier Transforms are equal. With $\hat{} = \mathcal{F}$,
$$
\begin{align}
{\hat b}(\omega) &amp;= \frac{1}{C_\psi} \int_0^{\infty} 
    {\hat f}(\omega) \sqrt{a} {\bar{\hat \psi}}(a\omega) \sqrt{a} {\hat\psi}(a\omega) \frac{da}{a^2} \tag{3a} \\
                 &amp;= \frac{{\hat f}(\omega)}{C_\psi} \int_0^{\infty} \tag{3b}
    |{\hat \psi}(a\omega)|^2 \frac{da}{a}
\end{align}
$$
With change of variables $\xi = a\omega$, we thus 'prove'
$$
{\hat b}(\omega) = {\hat f}(\omega) \left(\frac{1}{C_\psi} \int_0^\infty \frac{|{\hat \psi}(\xi)|^2}{\xi} d\xi \right) = {\hat f}(\omega). \tag{4}
$$
Interpretation:
It's tempting to attribute the $a^{-2}$ to the wavelet dilations, $\psi(t/a)$, as they manifest as rescalings of Frequency-domain components ($\cdot \sqrt{a}$), and while that's true, it's not the crux.
Recall when I said it's not just about energy? Consider what happens if the normalization is anything other than $a^{-2}$. To arrive at $(4)$, we relied on $a^1$ in denominator in $(3b)$, else there is no cancellation with $C_\psi$, so no reconstruction. Thus, one can conclude that $a^{-2}$ is set in retrospect - i.e. &quot;what do we make it to cancel $C_\psi$&quot;.
Note that we can't go the other way, i.e. redefine $C_\psi$ such that $a^{-2}$ is no longer needed - at least not entirely; per above derivation, it's clear that $\psi$ must somehow drop, thus $C_\psi$ must include it. The usage of exactly $C_\psi$ might be convention-based, since $C_\psi$'s applicability spans much outside the iCWT as the fundamentally defining criterion of wavelets (Farge, 1992).
So how to invert L1-normed CWT? Obvious from above.
Notes:

Proof followed Mallat, Theorem 4.4
Split integrals based on identity
Assumed $f \in \mathbf{L^2}(\mathbb{R})$ (square-integrable)
$\psi$ was also assumed real, and note the tilde is missing. I'm unsure why such a restriction was made, or whether Mallat implies that all real $\psi$ are self-dual, but the rescaling identities (e.g. $a \Rightarrow a^{-1}$ via $\mathcal{F}$) should hold regardless. Derivation 2 does not restrict $\psi$ to real.


Derivation 2: Resolution of Identity
For all $f, g \in \mathbf{L^2}(\mathbb{R})$ (real or complex), below holds:
$$
\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{da db}{a^2} W_f(a, b) \overline{ W_g (a, b)} = C_\psi \left&lt; f, g \right&gt; \tag{5}
$$
Begin proof by transforming to frequency domain:
$$
\begin{align}
(5) = \int \int \frac{da db}{a^2} &amp; \left[ 
      \int d\xi {\hat f}(\xi) |a|^{1/2} e^{-jb\xi} \overline{ {\hat \psi}(a\xi)} \right] \\ 
      \cdot &amp; \left[ \int d\xi' \overline{ {\hat g}(\xi')} |a|^{1/2} e^{j b\xi'} {\hat \psi}(a\xi')
      \right] \tag{6}
\end{align}
$$
This is similar to derivation 1, except $\psi$ replaced with another CWT of some different function $g$. In the brackets we have the Fourier transform of $F_a(\xi) = |a|^{1/2} {\hat f(\xi)} \overline{\hat \psi (a\xi)}$ times the Fourier transform of complex conjugate of $G_a(\xi) = |a|^{1/2} {\hat g(\xi)} \overline{\hat \psi (a\xi)}$. By the unitarity of the Fourier transform, we have
$$
\begin{align}
(6) &amp;= 2\pi \int \frac{da}{a^2} \int d\xi F_a(\xi) \overline{G_a(\xi)} \\
    &amp;= 2\pi \int \frac{da}{|a|} \int d\xi {\hat f(\xi)} \overline{ {\hat g}(\xi)} |{\hat \psi}(a\xi)|^2 \\
    &amp;= 2\pi \int d\xi {\hat f(\xi)} \overline{ {\hat g}(\xi)}  \int \frac{da}{|a|} |{\hat \psi}(a\xi)|^2 \ \ \ \ \text{(Fubini's theorem)}  \\
    &amp;= C_\psi \left&lt; f,g \right&gt;,
\end{align}
$$
where in last step a similar change of variables was used, $\zeta = a\xi$. $(6)$ can be read as
$$
f = C_\psi^{-1} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{da db}{a^2} W_f(a, b) \psi^{a, b}, \tag{7}
$$
quoting the source,

with convergence of the integral &quot;in the weak sense&quot;, i.e., taking the inner product of both sides of $(7)$ with any $g \in \mathbf{L^2}(\mathbb{R})$, and commuting the inner product with the integral over $a, b$ in the right-hand side, leads to a true formula.

This seems to say that the original derivation was for the inner product $\left&lt; f, g \right&gt;$, whereas $(7)$ reads only for $(6)$, which will have somewhat different properties. The &quot;original&quot; is recovered by convolving with a $g$. -- I'm also unsure what exactly &quot;in the weak sense&quot; means; this is the most I found (clarifications welcome).
Once again, it is evident that $a^{-2}$ was set in retrospect, depending on how the forward transform $W$ was normalized.
Source: Daubechies, Ten Lectures on Wavelets, Chapter 2