This is I guess a common problem that stems from the way we use the word "intensity". Normally, we would use intensity to mean something like the energy going through a unit area per unit time. In the case of the photoelectric effect, we instead generally mean the number of photons (per unit time), as it is this that decides how many photo-electrons are emitted, and directly determines the current.
Also note that there is no way we can have a "small" change in the number of photons, compared to unity. If you slightly increase the frequency of a monochromatic source (a phrase that typically means a source of fixed frequency) by an arbitrarily small amount, the energy-intensity will necessarily increase. Only when you get to the point where the increase in frequency can be compensated for by reducing the number of photons by exactly one is it possible to have the same energy-intensity as what you started out with. You can see that energy-intensity is somewhat more complicated to keep constant when you bring in photons.
Anyway, for such cases, all you have to do is look at the variation of saturation current with frequency for a fixed photon-number-intensity i.e. consider the frequency and photon number as independent variables.
EDIT: For the saturation current, we consider the case when all the electrons get there, and the number of these electrons (per unit time), which is the current, is given by the photon number. Making the electrons get there is the precise function of the anode voltage in these situations, and the saturation current appears with a very large voltage. Otherwise, the kinetic energies of the electrons is the basic reason behind the variation of current with the anode voltage.
You have made an assumption that the photo-electrons will be randomly distributed in all directions. This assumption defies the Law of Conservation of Energy and the Law of Conservation of Momentum.
The incoming radiation from source S has momentum and energy that must be conserved. When the incoming radiation hits the cathode surface, some of the energy and momentum is distributed throughout the metal lattice on average inward and at opposite to the normal angle with an energy of the work function. The tensile forces in the metal lattice structure ensure this uniform distribution and the energy is transformed into heat. The remaining energy must be reflected at some angle which conserves momentum and energy. This reflected energy and momentum is carried by the photo-electrons.
So, there are in fact no electrons emitted to the left in your diagram. In fact, photo tubes are generally designed with a curved cathode surface such that the photo-electrons are reflected and focused towards the anode.
It should also be noted that if photo-electrons were being scattered in all directions then we would have a build up of a net positive electric charge at the location of the creation of photo-electrons. Electrons would have to come in to replace this deficiency from somewhere, but this would be inconsistent. The electric field is either pointing towards or away from this point and the average electron drift will be determined by this electric field. It cannot be both at the same time.
In reality, the photo-electrons travel towards the anode and the cathode lead supplies an inrush of electrons to replace the holes created. And thus you have a current flowing through the tube, commutator and ammeter even with the potentiometer set to 0 V.
Best Answer
The important part of this experiment should be that there's a relatively sharp cutoff. "Classical" theory would say that the probility of electron ejection is proportional to the energy density, so a high-power long wavelength source would produce the same energy as a low-power, short wavelength source. In fact, photons below a certain energy level cannot eject electrons at all. This was one of the breakthrough experiments which showed that photons, or quantized packets of light energy, exist.