Young's Two-Slit Experiment Without Slits

This is not the same experiment, as here one tries to study interference between two photons emitted consecutively, rather than the wave nature of a single photon.

Of course, classically photons are electromagnetic waves, whose interference in two-slit geometric does not require quantum approach (just like the interference of waves in water).

Let me also note that the random count toss is not the same as quantum interference. The probability resulting from a random coin toss is $|psi_1|^2, |psi_2|^2$, rather than $|psi_1 + psi_2|^2$.

No, the fringes will not appear. Instead you will get the probability distribution you would naively expect from treating the photons as classical billiard balls with uncertain direction.

Your experiment should yield the same results as letting two separate experiments run --- one with just slit 1 and one with just slit 2 --- and then choosing at random which screen you look at each time.

The distribution you will see will then be $$P(text{photon lands at }x) = P(text{slit in position 1})P(text{photon from just slit 1 lands at }x)+P(text{slit in position 2})P(text{photon from just slit 2 lands at }x)$$ where $P(text{photon from just slit }atext{ lands at }x) = |psi_{slit, a}(x)|^2$.

Because all the terms on the right hand side are positive you'd only see dark spots on the screen at locations where both $P(text{photon from just slit 1 lands at }x)$ and $P(text{photon from just slit 1 lands at }x)$ are equal $0$. There's no interference that creates new dark spots.

If you don't have slits then your sources are free to generate photons that can travel pretty much anywhere. What is meant by the historical meaning of "interference" i.e. a sort of cancelling of energy, only gives you a superficial/outdated understanding. "Interference" is better replaced by acceptable paths or Feynman theory, certain paths resonate based on path length and photon energy. –