 Solution

The first term $g{A}_{\mu }{g}^{-1}$ lies in $\mathfrak{g}$ because the adjoint action preserves the Lie algebra. For the second, fix $x$ and consider the group curve $s\mapsto g(x+s{e}_{\mu })g(x{)}^{-1}$ through the identity. Its tangent at zero is $({\partial }_{\mu }g)g^{-1}$, so this is also in $\mathfrak{g}$. Since a Lie algebra is a vector space, $A_{\mu }^{\prime }\in \mathfrak{g}$.