Let $R$ be an arbitrary ring. The quotient $R/J(R)$, where $J(R)$ is the Jacobson radical, is the part of $R$ that acts nontrivially on the simple modules of $R$; said another way, it is the universal semiprimitive quotient of $R$. If $R$ is Artinian, then $R/J(R)$ is Artinian semiprimitive, which turns out to be equivalent to semisimple, so we can apply Artin-Wedderburn to $R/J(R)$ and learn something important about our original ring $R$ (namely everything we can learn from looking at simple modules).
It sounds a bit like by "are they group isomorphisms" you are really asking "are they only group isomorphisms and not ring isomorphisms?" Of course they are additive group isomorphisms, but they are actually more than that.
While the individual Hom groups aren't rings, the sum $\bigoplus_{1\leq i,j\leq k}\text{Hom}_R\left(M_i^{n_i},M_j^{n_j}\right)$ is a ring!
Perhaps the way to get comfortable with it is to imagine it as a "formal" ring of matrices with entries from the Hom groups. Then formal matrix multiplication composes them in a reasonable way such it is a ring. The matrix multiplication matches the composition in $\mathrm{End}_R(M)$ exactly through the map, so it is a ring homomorphism. This holds even if the $M_i$ lack their special properties (being pairwise nonisomorphic simple modules.)
In the passage from line (1) to (2), we simply eliminate a lot of superfluous terms in the direct sum and discover that the multiplication boils down to coordinatewise multiplication of a direct product of rings. That is what the pairwise nonisomorphic condition buys us.
So, the isomorphisms involved are ring isomorphism all the way through.
Here's another sort of example. Let $e$ be an idempotent in any ring with identity, and let $f=1-e$. Then one can show that $R\cong \begin{bmatrix}eRe&eRf\\fRe&fRf\end{bmatrix}$ as rings, where the multiplication on the right is formal matrix multiplication. Neither $eRf$ nor $fRe$ has to be a ring, but $eRe$ and $fRf$ are both rings.
You can look upon this as $R\cong\mathrm{End}_R(R)\cong \mathrm{End}_R(eR\oplus fR)=\mathrm{Hom}_R(eR\oplus fR,eR\oplus fR)\cong\begin{bmatrix}\mathrm{Hom}_R(eR,eR)&\mathrm{Hom}_R(fR,eR)\\\mathrm{Hom}_R(eR,fR)&\mathrm{Hom}_R(fR,fR)\end{bmatrix}$
The (group) isomorphism $\mathrm{Hom}_R(fR,eR)\cong eRf$ can be found in Lam's First course in noncommutative rings (Proposition 21.6.) It turns out to be a ring isomorphism if $e=f$.
Best Answer
You just need to look deeper at the theorem you mentioned.
The simplest example of how it works looks like this:
$End(M\times N)\cong \begin{bmatrix}Hom(M, M)& Hom(N,M)\\Hom(M,N)& Hom (N,N)\end{bmatrix}$ as rings where the multiplication is matrix multiplication. Take a look at how the elements from the various entries compose with each other and generalize it to your situation.
The final piece of the puzzle is to notice that when $M$ and $N$ are different Wedderburn components of $R$, there is no nonzero homomorphism between them. In our toy example above, if $Hom(M,N)=Hom(N,M)=\{0\}$, then you're looking at the diagonal matrix ring
$\begin{bmatrix}Hom(M, M)& 0\\0& Hom (N,N)\end{bmatrix}\cong Hom(M,M)\times Hom(N,N)=End(M)\times End(N)$, all as ring homomorphisms.