Yes it is. You are writing down the projection to the origin of the group through its character expansion, indeed a generalization of the Fourier transform, as you guessed.
A good book would cover it all, such as Vilenkin's classic, Ch I, §4. I'll use the physicist's informal seat- of-the-pants language and avoid mathematese, which probably got you stranded in the first place.
Your formula is the Fourier expansion for U(1), whose reps are all one dimensional, $e^{ir\theta}$, and the group-invariant Haar measure is trivial/flat, dθ ; so you get the δ(θ-0), having projected to the origin of the group.
The point is $\mathrm{tr}_r(U)\equiv \chi_r(U)$ is the character of the rep,
a class function, that is it is the same for the entire equivalent class containing U, by the similarity invariance of the trace.
The character of the identity, then, is just $\chi_r(\mathbb{1}) = \dim(r)$, the dimension of the rep. From the orthogonality of the characters outlined in the linked WP article, you then see you may represent U as a direct sum of this U in every rep, with its character in that rep as a coefficient, so, then, just like the Fourier expansion.
So you are writing the completeness relation of your group with the origin of the group as the destination.
Note however, in what you write, that the right hand side is a class function, so the sloppy l.h.s. must be interpreted as acting on class functions and dialing their angles to the origin, e.g. $\int dU \delta(U-\mathbb{1}) \chi_s (U)= \dim(s) $.
For SU(2), the simplest case, this is not that recondite. There is only one equivalence class--recall a similarity transformation can represent any group element as a rotation $R_3$ around some Euler axis arrived at by this similarity transformation: so you are just looking at $e^{i\theta J_3}$. The characters then are just Chebyshev polynomials of the second kind, a Dirichlet kernel,
$$
\chi_s(\theta)=\frac{\sin ((2s+1)\theta/2)}{\sin (\theta/2)},
$$
the obvious sum of the 2s +1 exponentials in the trace of $R_3$. The Haar measure is
$$
\int_0^{4\pi} d\theta \frac{\sin^2 (\theta/2)}{2\pi}.
$$
You are then summing
$$
(n+1)\frac{e^{i(n+1)\theta/2} - e^{-i(n+1)\theta/2}}{e^{i\theta/2} - e^{-i\theta/2}}
$$
over all positive integers n=2s+1. (Cheap and dirty evaluations may end up yielding 0/0. My favorite formal wisecrack might be defining $S(\theta)\equiv \sum_{n=1}^{\infty}e^{in\theta/2}$, evaluating $\partial_\theta (S(\theta)+S(-\theta))$, whence $-\partial_\theta \delta(\theta/2)/\sin(\theta/2)$, given the Fourier kernel of the periodic δ, the
Dirac comb,
but no need to take it too seriously!). Large ns will net you periodic δs, as per the Dirichlet kernel WP article, but the point is the smaller n terms conspire to interfere destructively.
Still, technicalities aside, the projection is bound to work out. Check out that
$$
\int_0^{4\pi} d\theta \frac{\sin^2 (\theta/2)}{2\pi} \frac{(-\partial_\theta \delta(\theta/2))}{2\sin(\theta/2)} ~ \chi_7 (\theta)= 15,
$$
after the integration by parts.
You may find practical examples of this technique in our paper here.
Note added in mathematese, since the OP appears interested in the structure of characters providing a complete orthonormal set for class functions. Fulton and Harris, p 440, summarize the Peter-Weyl theorem, which generalizes Fourier analysis to compact groups: "The characters of irreducible representations span a dense subspace of the space of continuous class functions." So the character completeness relation projects the equivalence class of U to the identity, an equivalence class of its own.
Here is a link to the stuff in Vilenkin & Klimyk 1991, not as good as the 68 classic, in my parochial view...
Finally, the WP section on D matrices' orthogonality illustrates completeness and the identity limit for SU(2), reviewed by Schwinger in eqn (3.96) of his monumental 1952 review.
