When you have a capacitor, current flows even though the "circuit" is not complete. This is because it's possible for electrons to bunch up - temporarily - in a conductor and generate a corresponding electric field.
That is what happens in an antenna. An antenna is really a combination of an inductor (a straight wire) and a capacitor (when you put a net charge on it, its potential will rise compared to the surroundings). It is this combination that allows an antenna to work - and to work best at a specific frequency, its resonant frequency.
Now when you drive a current into an antenna at resonance, you will find that the current is in phase with the voltage - that is when the circuit is called matched, and when it is possible to send power into the antenna. When you did not properly match your antenna (which is very easy to do!), then power will be reflected - this is a big problem with transmitters and it can actually fry the output circuitry. For this reason, transmitters have matching circuits between the power amplifier and the antenna, and a "VSWR meter" (Voltage Standing Wave Ratio meter) to measure the presence of reflection (which results in a partial standing wave).
There is a thing called the "radiation resistance" of an antenna that describes how the antenna behaves as a resistor at resonance - and how much power you can send into it for a given voltage. This follows directly from the usual power law:
$$P = \frac{V^2}{R}$$
Where $V$ is the voltage (RMS) and $R$ is the radiation resistance, $P$ is the power absorbed by the antenna (and emitted in the form of E/M radiation).
The key concept here is that the impedance of the antenna will be purely real at resonance. At that point, the capacitance and inductance of the antenna (intrinsic, or added) ensures resonance, and then significant currents can flow in phase with the voltage - which allows power to flow from the antenna into the radiated signal.
In electrostatics, the electric field in a wire loop is conservative (why?) so an electron which makes a full turn around wire loses as much energy as it gains. In other words the emf of the circuit is zero, if I understand correctly
Electrostatic field is conservative because if we move a charge from point A to B, the work done is irrespective of the path taken.
The easiest way to prove this is by taking a point $q$ charge, taking it away to some point and then bringing it back to initial point (without increasing its kinetic energy).
For electrostatic field:
$W_{ext}=-q\Delta V=-q(\oint_{closed-loop}\vec E \cdot {d\vec{l})}=-q(-\frac{\mathrm{d}\phi }{\mathrm{d} t})=0 \\ \Rightarrow W=0$
This implies that irrespective of the path taken, work done is always $0$ in an electrostatic field when charge is brought back to initial position since $\frac{d \phi}{dt} $ is always $0$.
For induced electric field:
$W_{ext}=-q\Delta V=-q(\oint_{closed-loop}\vec E \cdot {d\vec{l}})=-q(-\frac{\mathrm{d}\phi }{\mathrm{d} t}) \neq 0 \\ \Rightarrow W \neq 0$
Here $\frac{d\phi}{dt}$ depend on the path taken.
Hence electrostatic field is conservative, but induced electric field is not.
In otherwords the emf of the circuit is zero, if I understand correctly.
Since there is no battery, there is no EMF. In the above case, we are making the charge move around. So there is no need of EMF.
If you were asking about a circuit with a battery, then the EMF is provided by the battery.
However, apparently in a circuit with an induced emf, an electron which makes one full loop can gain energy. So as electrons continue to travel around the circuit, they gain more and more energy. In otherwords, the induced emf is due to an induced electric field in the wire which is nonconservative. Does this make any sense?
The work done in changing the magnetic field is transferred to the electrons in the conductor, which gives them energy to move around. This is a consequence of Law of Conservation of Energy.
Now, main question:
... However, doesn't this contradict our definition (*) of induced emf?**
No. Because it is not rigorous to use Faraday's Law here in the first place.
$\oint_{c}\vec E \cdot {d\vec{l}}=-\frac{d}{dt}\int_{s} \vec B \cdot {d\vec{l}} $
is the common version of Faradays's Law. It is rigorously correct only if $\vec E$ represents the electric field in the rest frame of each segment $d\vec l$ of the path of integration. This is definitely not true is the case of motional EMF.
It is necessary to note that only a time-varying magnetic field induces a circulating non-conservative electric field in the rest frame of the laboratory. A changing flux does not induce an electric field. Hence in case of motional EMF, no electric field is induced in the laboratory frame(where rod is in motion). Hence technically $\oint_{c}\vec E \cdot {d\vec{l}} = 0$ here.
A potential difference is maintained across the rod by the EMF and Lorentz force maintains that EMF. Using $-\frac{d}{dt}\int_{s} \vec B \cdot {d\vec{l}} $ here is just a convenient way of calculating the EMF.
Source: A Student's Guide to Maxwell's Equations -- Daniel Fleisch
Best Answer
Remember that the story we tell children about electric currents --- that energy in electric circuits is carried by moving electric charges --- is somewhere between an oversimplification and a fiction. Energy flow in electric circuits is carried by electromagnetic fields. That's true even for DC circuits, like the one below. You can think of the transmission line between the voltage source and the load as a capacitor, with an electric field $\color{red}{\vec E}$ pointing from the upper wire down to the lower, and as an inductor with a magnetic field $\color{green}{\vec H}$ pointing into the page in the loop and returning outside. The magnitude and direction of power flow is given by the Poynting vector $\color{blue}{\vec S} \propto \color{red}{\vec E}\times\color{green}{\vec H} $, which points out from the battery, down the transmission line, and into the resistor. If you do the hard, three-dimensional integral to find out how much of the power flow in this sort of circuit is described by the Poynting vector, you find ... all of it.
To find the distribution of charges on the ground plane, we have to satisfy the boundary conditions for the electromagnetic field at the surfaces of the conductors. Let's do a static charge on the conducting trace (for example: a constant-voltage signal to some FET with very large input impedance, so that it draws no current) first, to warm up. For a point charge above an infinite conducting plane at $z=0$, the method of image charges gives charge density on the ground plane
$$ \sigma = -\epsilon_0 \frac{\partial V}{\partial z} \Bigg|_{z=0} = \frac{-q a}{2 \pi \left(\rho^2 + a^2\right)^{3/2} } \tag 1 $$
where $\rho$ is the cylindrical distance from the point charge. Note that the total distance between a point $(\rho,\phi,0)$ on the ground plane and the point charge at $(0,\text{anything},a)$ is $r=(\rho^2+a^2)^{1/2}$. If $\theta$ is the angle between $\vec r$ and the vertical, we have $\sigma \propto r^{-3}a \propto r^{-2}\cos\theta$ for a point charge.
We got that result using the potential for a point charge (and its mirror image) of $V\propto r^{-1}$ for a distance $r$ from the charge. The potential due to a line charge goes like $V\propto \ln r$ for a (perpendicular) distance $r$ from the charge. Doing the same derivative as in (1) ought to then give us $\sigma \propto r^{-1}\cos\theta$, but I'll leave the details and $2\pi$s and so on to you.
So there's your derivation for a static line charge. How does the charge distribution on the ground plane change if there's a current? The answer is that it doesn't. The boundary condition on $\vec E$ at the ground plane is still that $\vec E$ must be normal to the plane. The current introduces everywhere a magnetic field $\vec B$ that's perpendicular to the direction of the line current. Consider the DC/zero-frequency case first: there's no $\partial\vec B/\partial t$ to change the electric field, so the charge density must be the same as in the static case. This rebuts the (reasonable) intuition you state in your question, that the ground-plane current should be magnetically repelled by the line current: doing so would set up a transverse electric field on the surface of the ground plane conductor.
For an AC current, you can think of the current trace as a line charge whose charge density varies sinusoidally along its length. The method of image charges suggests that the surface charge under the ground plane should still vary like $\sigma\propto r^{-1}\cos\theta$ for any "transverse slice" under the current trace, giving the same electric fields as if there were an anti-trace below the ground plane with the opposite charge distribution. That charge distribution does have nonzero $\vec E_\parallel$ at the ground plane, in the direction of the line current; those are the fields that, together with the changing $\vec B$, move the ground-plane charges in the direction parallel to the current flow. The job of proving that the ground-plane component of $\vec E$ transverse to the circuit trace vanishes everywhere I'll leave to you as well.
Here's another argument that the static and dynamic charge distributions transverse to the circuit trace should be the same, based on symmetry. (I think I learned about this problem from Griffiths's E&M textbook.) Imagine you have two infinite, parallel, opposite-sign line charges at rest relative to each other: they'll experience an electrostatic attraction, but no magnetic interaction. However, another experimenter walking past you sees the two line charges, in her boosted reference frame, sees the line charges as antiparallel currents, which have a magnetic repulsion in addition to their electrostatic attraction; the magnetic repulsion becomes stronger, in her reference frame, as she passed you more quickly and the apparent current becomes larger. It seems plausible that, if the other experimenter walks past you sufficiently rapidly, she might see the magnetic repulsion become larger than the electrostatic attraction. Then you would disagree about whether the two line charges were attracted or repelled from each other. It is instructive to compute the speed of the boost at which the magnetic and electric forces balance each other exactly --- though if you've studied special relativity you might be able to guess the answer.