Confusion - ideal transformer on load
I have a confusion about the fact that even though there is no
potential difference because E=U there is still a current flowing.

The induced voltage is in series with the inductance of the winding and, given that the induced voltage is exactly the same as the applied voltage, there is, in effect, zero volts across the primary inductance. Given that an inductor's inductive reactance is zero when 0 volts is applied to it, the current that flows is this: -
$$\color{red}{\boxed{\dfrac{0 \text{ volts}}{0 \text{ ohms}}}}$$
This is an indeterminate quantity and rightly so.
In other words, for an inductor, you cannot say that zero amps flows just because there is zero volts across it. Another example is the current in an inductor when the applied voltage is a sinewave: -

I've drawn purple circles on the above graph (taken from here). Those purple circles are placed to coincide with the applied voltage across the inductor equalling zero - what do you notice about the current? It's not zero of course.
Another example; take an inductor and applied (say) 1 volt to it then remove it from the supply and instantaneously short it out - does the current fall to zero or does it continue to flow. It's the latter of course and this serves to remind us that the current flowing in an inductor has little to do with the applied voltage currently present.
This however, is always true: -
$$\boxed{V = L\cdot\dfrac{di}{dt}}$$