In the article 'An introduction to Bundle Gerbes', Michael Murray says (on p.4) that :
It is a standard fact that isomorphism classes of $H$-bundles are in bijective correcpondance with $H^1(M,H)$. (Here $H$ is an abelian group)
Now we know that the set of isomorphism classes of $GL(n,\mathbb{R})$- principal bundles are in bijective correspondence with the set of isomorphism classes of $O(n)$-principal bundles.
For the case where $n=1$, both $GL(n=1,\mathbb{R})=\mathbb{R}^*$ and $O(n=1)=\mathbb{Z}/2\mathbb{Z}$ are abelian.
Can we therefore conclude that $H^1(M,\mathbb{R}^*) \simeq H^1(M,\mathbb{Z}/2\mathbb{Z})$ ?
I think I am missing something here.
Yes you can. In fact you can prove this directly, and this will show that $H^n(M,\mathbb{R}^*)=H^n(M,\mathbb{Z/2Z})$ for every $n>0$.
But to avoid confusion, I will change notation slightly. Indeed, $\mathbb{R}^*$ denotes here the sheaf of continuous function with values in $\mathbb{R}^*$. So I will write it $\mathcal{C}^*$, and I will write $\mathcal{C}$ for the sheaf of continuous function with values in $\mathbb{R}$. This is because $H^n(M,\mathbb{R})$ usually denote the cohomology of the sheaf of locally constant functions (and not all the continuous ones). For $\mathbb{Z/2Z}$, there are no ambiguities.
There is a short exact sequence $$0\longrightarrow\mathcal{C}\overset{\exp}\longrightarrow\mathcal{C}^*\overset{\operatorname{sign}}\longrightarrow\mathbb{Z/2Z}\longrightarrow 0$$ It gives a long exact sequence in cohomology $$ ...\longrightarrow H^n(M,\mathcal{C})\longrightarrow H^n(M,\mathcal{C}^*)\longrightarrow H^n(M,\mathbb{Z/2Z})\longrightarrow H^{n+1}(M,\mathcal{C})\longrightarrow ...$$ But $\mathcal{C}$ is a fine sheaf (it has partition of unity), so has vanishing cohomology for $n>0$. Hence, the sign morphism induces isomorphisms $H^n(M,\mathcal{C}^*)\overset\sim\longrightarrow H^n(M,\mathbb{Z/2Z})$ for $n>0$.