Here's how I tend to think about this: you want to show that $$\sigma_* \mathcal{O}_{\hat{X}}( (n-1) E + \sigma^* K_X) = \mathcal{O}_{X}(K_X).$$ This is actually more than you need.
So, by the projection formula and as you mentioned, that $K_{\hat{X}} = \sigma^*K_X\otimes \mathcal{O}((n-1)E)$, we only have to show that $$\sigma_* O_{\hat{X}}( (n-1) E ) = {O}_{X}.$$ The left side is rational functions which are regular on $X \setminus \{ x \}$, where $\sigma$ is an isomorphism, and which have poles of order at most $n-1$ at $E$.
Certainly we have the containment $\supseteq$ (functions that are regular on $X$ are regular on $\hat{X}$ also).
On the other hand set $U$ to be $X$ with the point $x$ removed. If $i : U \to X$ is the inclusion, then $i_* \mathcal{O}_{U}$ is all rational functions on $X$ that are regular except at $x \in X$. Certainly then $$O_X \subseteq \sigma_* O_{\hat{X}}( (n-1) E ) \subseteq i_* \mathcal{O}_U.$$
As you pointed out, by Hartog's theorem this composition is an isomorphism. Thus you get the statement you wanted as well.