This is a proof of the last claim in Sam's answer. It would be hard to read as a comment.

Since both groups in question naturally split into direct sums when $G$ is a direct sum, you can reduce to $G$ cyclic. There is an exact sequence:

$0 \to Hom(G^{\vee},\mathcal O_X^{\times}) \to \mathcal O_X^\times \to \mathcal O_X^\times \to 0$

with the map from $\mathcal O_X^\times$ to itself being the $n$th power map with $n$ the order of $G$. Taking cohomology gives, after taking cokernels and kernels:

$0\to H^0(X,\mathcal O_X^\times)/H_0(X,\mathcal O_X^\times)^n \to H^1(Hom(G^{\vee},\mathcal O_X^\times))\to Hom(G^\vee,H^1(\mathcal O_x^\times))\to 0$

$H^0(X,\mathcal O_X)^\times= \mathbb C^\times$, and $\mathbb C^\times$ is a divisible group, so the $n$th power subgroup is the whole group. This gives the isomorphism.

This is a proof of the last claim in Sam's answer. It would be hard to read as a comment.

Since both groups in question naturally split into direct sums when $G$ is a direct sum, you can reduce to $G$ cyclic. There is an exact sequence:

$0 \to \mathcal Hom(G^{\vee},\mathcal O_X^{\times}) \to \mathcal O_X^\times \to \mathcal O_X^\times \to 0$

with the map from $\mathcal O_X^\times$ to itself being the $n$th power map with $n$ the order of $G$. Taking cohomology gives, after taking cokernels and kernels:

$0\to H^0(X,\mathcal O_X^\times)/H_0(X,\mathcal O_X^\times)^n \to H^1(\mathcal Hom(G^{\vee},\mathcal O_X^\times))\to Hom(G^\vee,H^1(\mathcal O_x^\times))\to 0$

$H^0(X,\mathcal O_X)^\times= \mathbb C^\times$, and $\mathbb C^\times$ is a divisible group, so the $n$th power subgroup is the whole group. This gives the isomorphism.

This is a proof of the last claim in Sam's answer. It would be hard to read as a comment.

Since both groups in question naturally split into direct sums when $G$ is a direct sum, you can reduce to $G$ cyclic. There is an exact sequence:

$0 \to Hom(G^{\vee},\mathcal O_X^{\times}) \to \mathcal O_X^\times \to \mathcal O_X^\times \to 0$

with the map from $\mathcal O_X^\times$ to itself being the $n$th power map with $n$ the order of $G$. Taking cohomology gives, after taking cokernels and kernels:

$0\to H_0(X,\mathcal O_X^\times)/H_0(X,\mathcal O_X^\times)^n \to H^1(Hom(G^{\vee},\mathcal O_X^\times))\to Hom(G^\vee,H^1(\mathcal O_x^\times))\to 0$

$H_0(X,\mathcal O_X)^\times= \mathbb C^\times$, and $\mathbb C^\times$ is a divisible group, so the $n$th power subgroup is the whole group. This gives the isomorphism.

This is a proof of the last claim in Sam's answer. It would be hard to read as a comment.

Since both groups in question naturally split into direct sums when $G$ is a direct sum, you can reduce to $G$ cyclic. There is an exact sequence:

$0 \to Hom(G^{\vee},\mathcal O_X^{\times}) \to \mathcal O_X^\times \to \mathcal O_X^\times \to 0$

with the map from $\mathcal O_X^\times$ to itself being the $n$th power map with $n$ the order of $G$. Taking cohomology gives, after taking cokernels and kernels:

$0\to H^0(X,\mathcal O_X^\times)/H_0(X,\mathcal O_X^\times)^n \to H^1(Hom(G^{\vee},\mathcal O_X^\times))\to Hom(G^\vee,H^1(\mathcal O_x^\times))\to 0$

$H^0(X,\mathcal O_X)^\times= \mathbb C^\times$, and $\mathbb C^\times$ is a divisible group, so the $n$th power subgroup is the whole group. This gives the isomorphism.