The first option is correct. Your substitution function gives a residuum of 1/2.
Now, you need to transform it to the residuum of the original function.
Your function (around the pol z=1) can be written as a Laurent series: ##f(z)=\sum_{n=-1}^\infty a_n (z-1)^n ##. Note that all terms with n<-1 vanish because z=1 is a pol of first order. The residuum is defined as the first coefficient of the Laurent-series with negative index, i.e. a_{-1}.
Now, you found the residuum of the original function to be 1, i.e. a_{-1}=1.
If you do your substitution in the Laurentseries, namely (z-1)=2u, you will be left with a new Laurent series of a different function: ##g(u)=\sum_{n=-1}^\infty b_n (u-0)^n##, where ##b_n=a_n*2^n##. If you calculate the residuum of g in the usual way, you will get b_{-1}=1/2 (which you got).
Now you just have to calculate a_{-1} from that.
To explain your error: The residuum of the substituted function is not exactly the same as the residuum of the original function.