Completeness of an electron-electron-scattering in the Feynman diagram During an electron-electron-scattering in any case the electrons change their directions and by this undergo accelerations.

Shouldn’t the electrons emit photons in this time, loosing a part of its kinetic energy and slowing down a bit?
 A: 
Shouldn’t the electrons emit photons in this time...?

Yes. They do. Just as in classical physics, an accelerating (or scattering) electron in QFT emits EM radiation. If desired, this radiation can be described in terms of photons, although (as usual) that's not necessarily the most natural description.
To see this, consider the same Feynman diagram but with one or more external photon lines emanating from one or more of the electron lines. These amplitudes are non-zero, so this process does occur. In fact, when we consider higher-order terms in the small-coupling expansion (and these diagrams with extra photon lines are of higher order), we must include this effect in order to get a meaningful result. 
Weinberg's book (The Quantum Theory of Fields, Volume 1) has an entire chapter devoted to this subject (Chapter 13: "Infrared Effects"). Page 544 emphasizes that we must account for this effect even if we don't care about or don't detect the EM radiation:

...it is not really possible to measure the rate... for a reaction... involving definite numbers of photons and charged particles, because photons of very low energy can always escape undetected. What can be measured is the rate... for such a reaction to take place with no unobserved photon having an energy greater than some small quantity..., and with not more than some small total energy... going into any number of unobserved photons.

