Let's look at this another way for simplicity.
$\begin{eqnarray*} \sum_{k=-\infty}^\infty\left(\frac12\right)^{|k|} & = & \left(\frac12\right)^0+\sum_{k=-\infty}^{-1}\left(\frac12\right)^{|k|}+\sum_{k=1}^\infty\left(\frac12\right)^{|k|}\\ & = & 1+\sum_{k=-\infty}^{-1}\left(\frac12\right)^{-k}+\sum_{k=1}^\infty\left(\frac12\right)^k\\ & = & 1+2\sum_{k=1}^\infty\left(\frac12\right)^k. \end{eqnarray*}$
All we did was split it up a bit and reindex. Can you get the rest of the way from there?
Added: Here's a justification for what (to me) is the biggest leap in the originally-posted approach. A basic geometric series result (arguably, the most basic such result) is that for $|x|<1$, we have $\sum_{k=0}^\infty x^k=\frac1{1-x},$ so $\sum_{k=1}^\infty x^k=-x^0+\sum_{k=0}^\infty x^k=-1+\frac1{1-x}=-\frac{1-x}{1-x}+\frac1{1-x}=\frac{-(1-x)+1}{1-x}=\frac{x}{1-x}.$ Suppose that $|y|>1$, so if we put $x=\frac1y$--that is, $x=y^{-1}$--then $|x|=\left|\frac1y\right|=\frac1{|y|}<1$, so on the one hand, $\sum_{k=1}^\infty x^k=\sum_{k=1}^\infty\left(y^{-1}\right)^k=\sum_{k=1}^\infty y^{-k}\:\overset{j\,\mapsto\,k}=\:\sum_{j=1}^\infty y^{-j}\;\overset{k\,\mapsto\,-j}=\;\sum_{k=-\infty}^{-1}y^k,$ and on the other hand, $\sum_{k=1}^\infty x^k=\frac{x}{1-x}=\cfrac{\frac1y}{1-\frac1y}=\cfrac1{1-\frac1y}\cdot\frac1y=\cfrac1{\left(1-\frac1y\right)y}=\frac1{y-1}=-\frac1{1-y}.$ Thus, for $|y|>1$, we have $\sum_{k=-\infty}^{-1}y^k=-\frac1{1-y}.$ In particular, our "leap" from $\sum\limits_{k=-\infty}^{-1}2^k$ to $-\frac1{1-2}$ is simply the above result, in the special case $y=2$.