The "Maz identity" states: $$ \int_0^\infty f(x)g(x)\mathrm{d}x = \int_0^\infty \mathcal{L}\{f\}(u)\mathcal{L}^{-1}\{g\}(u)\mathrm{d}u, $$ where $\mathcal{L}$ is the Laplace transform.
I came across this identity when trying to find the Mellin transform of $\sin(x)$. The theorem turns out to be very useful, but I could not find any reference for a proof of this identity. The only references on MSE are this and this, but neither provides a derivation. This identity also appears in a recent IG post by owenmmth, for those who are interested.
PS: Although not important to the question, but the Mellin transform is defined as $$ \{\mathcal{Mf}\}(s) = \int_0^\infty x^{s-1}f(x)\mathrm{d}x .$$
As Zima pointed out in the comment, Wikipedia's page on the Laplace transform gives a nice proof, so I thought I would answer my own question for others who happen to search for this.
Start by showing that
$$ \int_0^\infty h(x) \{\mathcal{L}g\}(x)\mathrm{d}x = \int_0^\infty \{\mathcal{L}h\}(x)g(x)\mathrm{d}x. $$
This is obvious from the definition of the Laplace transform and an application of Fubini's theorem:
$$ \begin{align} \int_0^\infty h(x) \{\mathcal{L}g\}(x)\mathrm{d}x &= \int_0^\infty h(x) \int_0^\infty g(u)e^{-ux}\mathrm{d}u \mathrm{d}x \\ &= \int_0^\infty g(u) \int_0^\infty h(x)e^{-ux}\mathrm{d}x \mathrm{d}u \\ &= \int_0^\infty \{\mathcal{L}h\}(u)g(u)\mathrm{d}u \\ &= \int_0^\infty \{\mathcal{L}h\}(x)g(x)\mathrm{d}x . \end{align} $$
Now make the substitution $h(x) = \{\mathcal{L}^{-1}f\}(x)$, so that $\{\mathcal{L}h\}(x) = f(x)$:
$$ \int_0^\infty \{\mathcal{L}^{-1}f\}(x) \{\mathcal{L}g\}(x)\mathrm{d}x = \int_0^\infty f(x)g(x)\mathrm{d}x. $$
This proves the useful Laplace transform identity.