Who has to buy the beer?

I don't understand the method you are attempting but here is a solution to the problem. Consider all sequences $$x_0,x_1,\ldots,x_n$$ with $n$ steps, of the type you are considering: that is $$x_k\in\{1,\ldots,115\},\ x_0=1,\ x_k\ne x_{k-1},\ x_k\ne x_{k-2}.$$ We wish to find $P(x_n=1)$ in the case $n=2009$.

I will assume that people do not "target" the leader as suggested in other comments but that they choose at random from all available possibilities.

Let $p_n$ be the probability of obtaining $x_n\ne1$ and $x_{n-1}\ne1$ in such a sequence. The answer to your problem is $$\frac{p_{2008}}{113}\ .$$ Now for $n\ge4$ we have $$\eqalign{p_n &=P(x_n\ne1,\,x_{n-1}\ne1,\,x_{n-2}\ne1) +P(x_n\ne1,\,x_{n-1}\ne1,\,x_{n-2}=1)\cr &=\frac{112}{113}p_{n-1}+P(x_{n-1}\ne1,\,x_{n-2}=1)\cr &=\frac{112}{113}p_{n-1}+P(x_{n-2}=1,\,x_{n-3}\ne1,\,x_{n-4}\ne1)\cr &=\frac{112}{113}p_{n-1}+\frac{1}{113}p_{n-3}\ .\cr}$$ This recurrence relation, with suitable initial conditions, can be solved in the usual way. The characteristic equation is cubic: fortunately, $1$ is a root; unfortunately, the other two are messy complex surds. If my algebra is correct the answer is $$p_n=\frac1{(1-\alpha)(1-\beta)(\alpha-\beta)}\bigl[(\alpha-\beta)-(1-\beta)\alpha^n+(1-\alpha)\beta^n\bigr]$$ where $$\alpha=\frac{-1+i\sqrt{451}}{226}\ ,\quad \beta=\frac{-1-i\sqrt{451}}{226}\ .$$ Evaluation by Maple gives the answer $$\frac{p_{2008}}{113}=0.008695652174$$ which is very close to $\frac1{115}$. This is to be expected since, as pointed out in other comments, any initial effects reducing (or increasing) the chance of the leader buying the beers should have almost disappeared by the $2009$th step.

Intuitively, after just a few messages we will forget where we started. In that case, each person has equal chance to receive the $n$th message, so the chance the leader buys the beer is $1/115$. To explicitly demonstrate a path we can go around a cycle of all $115$ people $17$ times, go through $53$ points of the next cycle and back to the lead.

On the other hand, it is reasonable to suppose that everybody on the team wants the lead to buy the beer. When the lead sends a text, the next person can't return it so sends it to someone else. That person can tag the lead, so the lead receives every other message. As $2008$ is even, the lead will get the $2008$th one and gets to pick the victim with the $2009$th.