What determines photoelectric yield? Is there any difference between the photoelectric yield of different metals apart from the threshold wavelength?
To be more clear: Will metals with the same work function emit the same amount of electrons per incident photon?
 A: The photoelectric work function is primarily a surface effect, and for a given metal will vary significantly by crystal face. Note the variations given for silver, with the lowest, 4.26 eV, being from the poly-crystalline form.
Modelling of efficiency is complicated; from a macroscopic viewpoint one has the skin depth of the metal by wavelength, which provides the effective rate of photon absorption,  but this doesn't tell you which electrons will be scattered towards the surface. In a naive model only those electrons which only undergo the first scattering event can make it out, because they lose some energy with each interaction,  slowly becoming thermalized.
Because the situation is so complex, you should not be surprised that the quantum efficiency varies significantly by metal, but less so by slight variations of the photon energy, as long as it exceeds the work function.  Cesium is orders of magnitude more efficient than gold. One photo-electron per $10^5$ or $10^6$ photons is typical for gold and silver.  I have some measurements in my dissertation, for ultrafast laser pulses at 266 nm, and 200 nm thick polycrystalline gold in transmission;  I'll look them up later.
With ultrafast laser pulses, where the photon density is very high, muli-photon absorption can occur, which is a nonlinear effect. So three 780 nm photons, each about 1.5 eV, can be absorbed by gold within a few femtoseconds,  emitting a single electron with a work function close to 4.5 eV.
