Generally, it isn't the case that the (net, total) magnetic flux threading the conducing loop is constant.
Note that the quote uses the word opposes which you have evidently taken to mean nullify. But in fact, if the solution requires an emf, then there must be a non-constant flux threading the loop.
When the conductive material forming the loop has non-zero resistivity, the solution must satisfy
$$\mathscr E = -\frac{\mathrm d\Phi}{\mathrm dt}$$
and
$$\mathscr E = R i$$
where $R$ is the resistance 'round the loop and $i$ is the current.
But
$$\Phi = \Phi_\textrm{ext} + \Phi_i = \Phi_\textrm{ext} + L i$$
where $L$ is the inductance of the conducting loop.
Combining the above yields
$$R i + L \frac{\mathrm di}{\mathrm dt} = -\frac{\mathrm d\Phi_\textrm{ext}}{\mathrm dt}$$
For example, consider the case that $\Phi_\textrm{ext} = \Phi_0$ is constant then the solution is
$$i(t) = I_0e^{-\frac{R}{L}t}$$
$$\Phi(t) = LI_0e^{-\frac{R}{L}t} + \Phi_0$$
$$\mathscr E(t) = -\frac{\mathrm d\Phi}{\mathrm dt} = -\left(-\frac{R}{L} \right)LI_0e^{-\frac{R}{L}t} =Ri(t)$$
So, this is all consistent and notice that the flux is not generally constant though the external flux is.
Now, consider the case that $\Phi_\textrm{ext}$ is increasing linearly with time
$$\Phi_\textrm{ext} = \Phi_0 + \phi t$$
then the particular solution is
$$i(t) = -\frac{\phi}{R}$$
$$\Phi(t) = \left(\Phi_0 -\frac{L}{R}\phi\right) + \phi t$$
$$\mathscr E = -\phi = Ri $$
Again, this is consistent and the flux is not generally constant. However, note that this result depends on $R$ being non-zero.
For the $R = 0$ case, we see that
$$L \frac{\mathrm di}{\mathrm dt} = -\frac{\mathrm d\Phi_\textrm{ext}}{\mathrm dt}$$
or
$$\Phi_i = - \Phi_\textrm{ext} + \mathrm{constant}$$
$$\mathscr E = -\frac{\mathrm d\Phi}{\mathrm dt} = 0$$
thus we conclude that, for a perfectly conducting loop, the magnetic flux threading the loop is constant.
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
You are correct.
There are a number of ways of looking at the situation.
Assume that there is a magnetic field $\vec B$ into the screen and the whole circumference of the loop is moving inwards at a speed $\vec v$.
The direction of the induced current can be deduced from the right hand grip rule knowing that Lenz's law predicts that the magnetic field produced by the induced current will be in such a direction as to reduce the change producing it which is a reduction in the magnetic flux through the loop.
The induced magnetic field is into the screen and the induced current is clockwise.
A charge $q$ within the loop moving inwards at a velocity $\vec v$ experiences a Lorentz force $q \vec v \times \vec B$ in a direction as shown in the diagram.
This results in the direction of the induced (conventional - as if the mobile charge carriers were positive) current being clockwise.
Now you have an induced current-carrying conductor moving inward at a speed $v$ in a magnetic field $\vec B$.
There is therefore a force on the conductor radially outwards.
To maintain the radially inward constant speed motion of the loop an external force needs to be applied radially inwards and this external force does work as the4r loop moves inwards.
The work done by the external force results in the generation of the induced current which when passing through the loop which has resistance produces a heating effect.
This is the law of conservation of energy in action with the work done by the external force ultimately producing heat.