Log-linearisation Euler Consumption Equation
To simplify matters, let's call the right-hand side of your starting equation $X_t$ Then, I start just like you did with

$1 = X_t$

The difference between your solution and Gali's is that you took the Taylor expansion around

$\log(1)=0=log(X_t)$  which implies that also the steady state equals 0, so we can simply subtract it to get log-differences,

whereas Gali used

$1 = \exp(\log(X_t))$ with steady state $1=\exp(\log(X^*))$

Let's define $x_t=\log(X_t)$.

By definition of the Taylor expansion this gives

$T(1) = \exp(x^*)+\exp(x^*)(x_t-x^*)$ 

Note that the exponential function is identical to its derivative. We know that $\exp(x^*)=1$ from above, so we can simplify to 

$T(1)=1+(x_t-x^*)$ 

At this point, you are correct that the $\rho$ cancel out. Instead of defining $\hat{x}_t=x_t-x^*$ but Gali chooses to replace $\rho=\pi +\sigma \gamma-i$, which is the steady state condition, which is why $\rho$ stays on the paper. 

Note, however, that the "approximation error" from taking log differences instead of Gali's approach is tiny, i.e. for $\beta=0.99$ it is only $-\log(0.99)\approx 0.01$

Have a look at appendix 2.1 of Gali's book! It is quite complicated, but together with this post, I hope you get it!