Consider the extension operator: $$ Eg(x)=\int_S g(\xi)e^{2\pi i x\cdot \xi}d\sigma(\xi). $$ For simplicity we consider the 2D-case, where $S$ is the paraboloid $\xi\mapsto \xi^2$, $\xi\in [-1,1]$. (We will frequently use an abuse of notation: $g(\xi_1)=g(\xi_1,\xi_1^2)=g(\xi)$, where $\xi_1\in [-1,1]$ and $\xi\in S$.)
We know $EE^*f=f*\widehat {d\sigma}$. But what is $E^*E$? I claim that $E^*E=I$, and here is my formal computation. \begin{align*} Eg(x) &=\int_{-1}^1 g(\xi_1)e^{2\pi i (x_1,x_2)\cdot (\xi_1,\xi_1^2)}J(\xi_1)d\xi_1\\ &=\int_{-1}^1 \int_{\mathbb R} \delta_0(\xi_1^2-\xi_2)g(\xi_1)e^{2\pi i (x_1,x_2)\cdot (\xi_1,\xi_2)}J(\xi_1)d\xi_2d\xi_1. \end{align*} Hence, formally, $$ \widehat {Eg}(\xi_1,\xi_2)=g(\xi_1)J(\xi_1)\delta_0(\xi^2_1-\xi_2) $$
Thus $\widehat {Eg}$ is a distribution in $\mathbb R^2$ supported on the surface $S$, in the sense that if $h\in C_c^\infty(\mathbb R^2)$, then \begin{align*} \widehat {Eg}(h) &=\int_{\mathbb R^2} g(\xi_1)J(\xi_1)\delta_0(\xi^2_1-\xi_2)h(\xi_1,\xi_2)d\xi_2d\xi_1\\ &=\int_{\mathbb R} g(\xi_1) J(\xi_1)h(\xi_1,\xi_1^2)d\xi_1\\ &=\int_{S} g(\xi)h(\xi)d\sigma(\xi). \end{align*} Then $E^*Eg$ is the restriction of $\widehat {Eg}$ to $S$, so $E^*Eg$ is exactly the function $g$ defined on $S$. In other words, $E^*E=I$.
The place I am not sure of is the last line, because I do not know in general, how I could restrict a distribution to a submanifold.