I can't tell if you want a reference to a proof or just a proof, but here's a proof.
By multiplying by a suitable power of the determinant we may assume that such a homomorphism extends to a polynomial monoid homomorphism $\mathcal{M}_n(k) \to k$, and since it is polynomial it extends to a homomorphism $\phi : \mathcal{M}_n(\bar{k}) \to \bar{k}$. Since $\bar{k}$ is abelian, $\phi$ is conjugation-invariant. Restricted to diagonal matrices $\phi$ must be a symmetric polynomial in their diagonal entries, and since the diagonalizable matrices are Zariski dense in $\mathcal{M}_n(\bar{k})$, $\phi$ must be a symmetric polynomial in the eigenvalues (hence a polynomial in the coefficients of the characteristic polynomial) identically. By scaling entries of elements of $\mathcal{M}_n(\bar{k})$, $\phi$ must be a homogeneous polynomial.
Assume WLOG that $\phi$ is not divisible by the determinant. If $\phi$ has degree $0$ as a homogeneous polynomial, then it must be the trivial homomorphism. Otherwise, $\phi$ is a homogeneous polynomial of positive degree, so in particular $\phi(0) = 0$. Now, if $D_1$ is a diagonal matrix with a zero entry, then we can find conjugates $D_2, ... D_n$ of $D_1$ which are also diagonal but with the zero entry at every possible other location. It follows that $D_1 ... D_n = 0$, so $\phi(D_1 ... D_n) = \phi(D_1)^n = 0$, so $\phi(D_1) = 0$. Hence $\phi(M)$ vanishes if any of the eigenvalues of $M$ vanish, so $\phi$ is divisible by the determinant; contradiction.
Edit: Your proposed extension is false. Take $G = \text{GL}_1 \times \text{GL}_1$ and embed using different powers for each factor...
