In molecular orbital theory, why does s-p mixing in the dinitrogen molecule not effect the 1σᵤ orbital?
Rob, if you poke around on the internet you can find many sites that show diagrams for how the $\ce{N2}$ molecular orbitals are formed from two sets of nitrogen ($\ce{N1}$) atomic orbitals.  Many of the diagrams are different one from another - that's a tip off that s-p mixing is not something that can always be predicted in advance.  With that understanding, let's step through your diagram and questions.

It might be useful to take a look at the $\ce{N2}$ MO diagram here.  It's just like your diagram. but they break down the steps and it might be a bit less confusing.

In any case we have a nitrogen atom with 5 valence electrons, 2 in the 2s atomic orbital and 1 each (total of 3) in the 3 degenerate p orbitals.  If we move a second nitrogen atom into the vicinity, these atomic orbitals will hybridize in order to start the bond formation process.  

$\ce{N2}$ is analogous to acetylene, they both have a triple bond that is made up of 1 sigma bond and 2 pi bonds.  In acetylene - and nitrogen - the sigma bond is $\ce{sp}$ hybridized.  We form this bond by


Mixing the s and  a p atomic orbitals (the p orbital whose lobe is linear with the s orbital [see the diagrams in the link I supplied above, they call it the $\ce{p_{z}}$ orbital]) on one of the nitrogens creates a bonding and antibonding $\ce{sp}$ orbital on that nitrogen atom.  
Same thing happens on the other nitrogen atom
As the 2 nitrogen atoms begin to interact and form bonds, the 2 $\ce{sp}$ orbitals on each nitrogen interact (split further).
The bonding $\ce{sp}$ orbital on each nitrogen interacts to form a lower energy and higher energy pair of bonding molecular orbitals (the $\ce{1\sigma _{g}}$ and $\ce{1\sigma _{u}}$ orbitals in your diagram - BTW, there should be a dashed line from the p orbitals to the $\ce{1\sigma _{g}}$ orbital - it is missing in your diagram, maybe that has led to some of your confusion).
The antibonding $\ce{sp}$ orbital on each nitrogen also interact in the same manner to create 2 antibonding sigma molecular orbitals ($\ce{2\sigma _{g}}$ and $\ce{2\sigma _{u}}$)


Now for pi bonds,


The remaining two p atomic orbitals on each nitrogen are orthogonal to the s and $\ce{p_{z}}$ orbital - they can't mix with those orbitals.
These atomic p orbitals $\ce{p_{x}}$ and $\ce{p_{y}}$ can only mix with their like counterpart on the other nitrogen atom.  This gives rise to a bonding and antibonding $\ce{p_{x}}$ pi molecular orbital and a bonding and antibonding $\ce{p_{y}}$ pi molecular orbital


With that background, let's now look at your questions


  s-p mixing in N 2   causes the 1σ g   orbital to decrease in energy,
  the 2σ g   orbital to increase in energy (to higher than the 1π u  )
  yet the 1σ u   is left unchanged by the interaction. Why is this?


I'd say it a bit differently as I did above. Mixing of the 2 bonding $\ce{sp}$ orbitals, one on each nitrogen atom, causes them to split (one is raised and one is lowered).  The same thing happens to the antibonding $\ce{sp}$ orbital on each nitrogen - one is raised, one is lowered.


  Clearly they are the s and p orbitals but is this within the atom or
  is the s orbital on one atom interacting with the p on the other?


Hopefully I answered this above.  I tried to distinguish between atomic and molecular orbitals.  If I haven't been clear, post back.


  Some clarification as to why the 1σ g   orbital to decreases in
  energy, the 2σ g   orbital to increases in energy would also be much
  appreciated.


Hopefully what I wrote above clarified this as well, let me know if it didn't.