4-momentum is conserved in any interaction, regardless of how you write it down. However, if the Feynman diagrams are expressed in position space, that's not immediately apparent, because you don't assign definite momenta to each line. In momentum space the lines all have momenta assigned to them, so you can add them up at each vertex and see that they always sum to zero, but in position space you have to take a Fourier transform to see that.
So I guess the answer to your question is that four-momentum conservation at vertices doesn't make sense in the position representation, because talking about momentum at all doesn't make sense in the position representation. But the math will work out either way to show that the overall interaction conserves momentum.