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 $X$).  

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.