Is there a simple example that shows that the path integral adds to zero for diagrams in which momentum or energy are not conserved? I know that if the result of the Hamiltonian or Lagrangian is constant through time then it means Energy is conserved by the process.
I don't immediately see that if the Hamiltonian or Lagrangian were not constant through time, the path integral will sum to zero.
To take a stab at it, ignoring other conservation laws: If the frequencies of a destroyed electron and created photon are equal, the resulting Hamiltonian or Lagrangian will be constant in time because the coefficients of wavelength or frequency match and produce a constant result.  If the energies do not match then it will clearly not be constant in time.  I do not clearly see the next step, is this sufficient to argue that there is no one constant contribution through time therefore many processes cancel?
 A: It is the same mechanism that is responsible for the rejections of paths that do not obey the physical laws. Feynman explains it in his book QED. The notion of "not constant in time" means that it oscillates. The amplitude of the oscillation is more or less constant. So, when you add up (integrate over) the function values as it oscillates, the result becomes close to zero after each cycle.
In the case of the non-conservation of momentum or energy, these oscillations come from the mismatch in the frequency or wave vector in the exponents associated with the quantization of the fields. The bigger the mismatch, the faster these oscillations. Therefore they quickly sum to zero.
This is a general mechanism. To see how it works in a specific case, one can work through an example of that case.
A: If you're looking for something general, beyond perturbation theory, I can only refer you to Weinberg's QFT volume 1. There you will find non-perturbative arguments about the presence of delta functions enforcing energy and momentum conservation. If you are content with perturbation theory as the question seems to imply, then let me point you in the right direction.
Every (momentum space) diagram comes attached to a delta function which imposes energy and momentum conservation of that diagram. Indeed, things are actually stronger than this as every single vertex comes with a 4-momentum-conserving delta function. Given this later fact, I will leave it to you to convince yourself that conservation at every vertex implies overall conservation for the diagram.
So, how can we understand where these delta functions come from at every vertex? The Feynman rules usually instruct us to just impose this fact manually, which is why we are left with integrations over only loop momenta and not over the momenta of every edge in the diagram: we in principle integrate over the momenta of every edge in the graph, but the momentum-conserving delta functions at each vertex kill all the integrations but the loop momenta (can see that this counting works out by looking at the Euler relation between edges, vertices, and number of loops).
The problem at hand is therefore reduced to the question: from whence does the Feynman rule about momentum conservation at each vertex come? For that matter, we could ask where any of the Feynman rules come from. They aren't magic after all, they are just instructing us on a systematic way to construct the generating function (path integral with sources included).
So, if you've never done it before, I recommend you spend some time deriving for yourself why the Feynman rules are what they are. It's an instructive exercise in and of itself, if a bit of a slog at times. As a hint, the particular step at which you will see momentum conservation appear on the interaction vertices is when you convert the fields in the action to their Fourier transforms.
