In an atom, the electron is not just spread out evenly and motionless around the nucleus. The electron is still moving, however, it is moving in a special way such that the wave that it forms around the nucleus keeps the shape of the orbital. In some sense, the orbital is constantly rotating.
To understand precisely what is happening lets calculate some observables. Consider the Hydrogen $1s$ state which is described by
\begin{equation}
\psi _{ 1,0,0} = R _1 (r) Y _0 ^0 = R _{1,0} (r) \frac{1}{ \sqrt{ 4\pi } }
\end{equation}
where $ R _{1,0} \equiv 2 a _0 ^{ - 3/2} e ^{ - r / a _0 } $ is some function of only distance from the origin and is irrelevant for this discussion and the wavefunction is denoted by the quantum numbers, $n$, $ \ell $, and $ m $, $ \psi _{ n , \ell , m } $. The expectation value of momentum in the angular directions are both zero,
\begin{equation}
\int \,d^3r \psi _{ 1,0,0 } ^\ast p _\phi \psi _{ 1,0,0 } = \int \,d^3r \psi _{ 1,0,0 } ^\ast p _\theta \psi _{ 1,0,0 } = 0
\end{equation}
where $ p _\phi \equiv - i \frac{1}{ r } \frac{ \partial }{ \partial \phi } $ and $ p _\theta \equiv \frac{1}{ r \sin \theta } \frac{ \partial }{ \partial \theta } $.
However this is not the case for the $ 2P _z $ state ($ \ell = 1, m = 1 $) for example. Here we have,
\begin{align}
\left\langle p _\phi \right\rangle & = - i \int \,d^3r \frac{1}{ r}\psi _{ 1,1,1} ^\ast \frac{ \partial }{ \partial \phi }\psi _{ 1,1,1} \\
& = - i \int d r r R _{2,1} (r) ^\ast R _{ 2,1} (r) \int d \phi ( - i ) \sqrt{ \frac{ 3 }{ 8\pi }} \int d \theta \sin ^3 \theta \\
& = - \left( \int d r R _{2,1} (r) ^\ast R _{2,1} (r) \right) \sqrt{ \frac{ 3 }{ 8\pi }} 2\pi \frac{ 4 }{ 3} \\
& \neq 0
\end{align}
where $ R _{2 1} (r) \equiv \frac{1}{ \sqrt{3} } ( 2 a _0 ) ^{ - 3/2} \frac{ r }{ a _0 } e ^{ - r / 2 a _0 } $ (again the particular form is irrelevant for our discussion, the important point being that its integral is not zero). Thus there is momentum moving in the $ \hat{\phi} $ direction. The electron is certainly spread out in a "dumbell" shape, but the "dumbell" isn't staying still. Its constantly rotating around in space instead.
Note that this is distinct from the spin of an electron which does not involve any movement in real space, but is instead an intrinsic property of a particle.
The orbital wavefunctions of the hydrogen atom, which obey the eigenvalue equation
$$
\left[-\frac{1}{2\mu}\nabla^2-\frac{e^2}{r^2}\right]\psi_{nlm}=E_{nl}\psi_{nlm},
$$
are functions of the separation vector $\mathbf r$ which points from the proton towards the electron. This is a standard trick in the two-body problem and it is done in both the classical and the quantum versions to factor away the motion of the bigger body (which is close to the centre of mass) and leave an effective one-body problem which is easier to treat.
This means that the orbital angular momentum, with total angular momentum number $l$, is in fact the combined angular momentum of the electron and the proton about their centre of mass. In essence, the proton partakes in part of the orbital motion and takes out some of the angular momentum from the electron. (Note, though, that this is classical language and it explicitly does not hold for the hydrogen atom, where the angular momentum of the motion is essentially indivisible.)
This raises an apparent paradox, which is resolved through the fact that the separation vector obeys dynamics through the reduced mass $\mu=1/\left(\tfrac{1}{m_e}+\tfrac{1}{m_p}\right)\approx\left(1-O\left({m_e\over m_p}\right)\right)m_e\lesssim m_e$ of the system, and this is slightly smaller than the electron mass. This slightly enlarges the orbital radius of the electron (since the Bohr radius is inversely proportional to the mass). The velocity stays constant (at $\alpha c$), which means that the angular momentum $L\sim \mu r v$ stays constant as well.
That said, the proton does have spin angular momentum of its own, but this couples weakly to the electronic motion. This coupling is via the same spin-orbit couplings as the electron, but its much higher moment of inertia means that the relevant energies are much smaller, as are the corresponding hyperfine splittings in the spectrum.
Best Answer
It is a misconception to say that
In QM the notion of "circling round the nucleus" does indeed fail to make sense, but this is not why electrons don't radiate EM waves. Instead, an atom in its ground state will not radiate because its state space simply does not contain any states with lower energy.
In essence, this is due to the uncertainty principle. If you have a hydrogen atom and you try to compress the 'electron orbit' any further, then the electron's position uncertainty will decrease, and this will drive a corresponding increase in the momentum uncertainty. This momentum uncertainty is proportional to $p^2$ and thus to the kinetic energy. This means that a decrease in the mean radius will decrease the potential energy but increase the minimum allowed kinetic energy, so there is a trade-off involved. For the ground state, this trade-off is optimal, and you cannot increase or decrease the orbit's spatial extent without increasing the energy.
On another track, you ask
The answer to that is that you need to stop thinking about the electron "circling" anything. The only relevant physical quantity is the electron's wavefunction. The electron is said to have orbital angular momentum if its wavefunction has significant changes in phase over paths which go around the nucleus. That's it.
Finally, in terms of
you are severely underestimating the extent to which angular momentum arises as the algebra associated with rotational symmetry within classical physics. Indeed, the algebra is exactly the same; you only replace commutators $i[·,·]$ with Poisson brackets $\{·,·\}$. Angular momentum is as related to rotational symmetry in classical physics as it is in quantum mechanics.