Two methods for calculating the average roll on 2d6 (including a single re-roll of 1 on one die) give different results - which is correct?

The first method is incorrect because it results in $\frac 16$ chance of rerolling instead of $\frac {11}{36}$. It is as if you roll one die, which you reroll if it comes up $1$, then roll a second which you cannot reroll. Th second looks right but I did not work through the details.

The first method is definitely not correct as RossMillikan explains.

Your second method is a bit involved, but let me do what I would do, and see if I get the same answer.

OK, I would start with the $36$ possible outcomes for the initial throw of two dice, and then take into account the reroll of those where at least one $1$ appears.

So: for those rolls with no $1$ you get $\frac{1}{36}*(1*4+2*5+3*6+4*7+5*8+4*9+3*10+2*11+1*12)$

For the rolls with a single $1$ you get $\frac{1}{36}*(2*(5.5+6.5+7.5+8.5+9.5)$ (e.g. if you initially throw a $2$ and a $1$, you can expect to get $2+3.5=5.5$)

And getting two $1$'s in the initial throw gives you $\frac{1}{36}*(1+3.5)$

All of this tgether totals: $\frac{1}{36}*(200+75+4.5)=\frac{1}{36}*279.5$

... which I can confirm coincides exactly with your second method's finding, which works out to $\frac{1677}{216}$