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 $e=f$.
You can just repeat step by step the proof for rings, by observing that, if $S$ is a simple $R$-module, then $D=\operatorname{End}_R(S)$ is a $K$-algebra in a natural way and the same for the endomorphism ring of $S$ as a $D$-module.
By the way, you don't need $K$ to be a field; just a commutative ring suffices.
Anyway, you can just apply the first two parts. Since $A$ is a $K$-algebra to begin with, also each $\operatorname{Mat}_{m_i}(D_{i})$ is a $K$-algebra; then $D_i$ is a $K$-algebra. The way you build these structures of $K$-algebras ensure the initial isomorphism is an isomorphism of $K$-algebras.
Let's see why. First, assume $R\times S$ is a $K$-algebra; then also $R\times S/{0}\times S$ is a $K$-algebra. We use this structure and the obvious ring isomorphism $R\times S/{0}\times S\to R$ to endow $R$ with a structure of $K$-algebra. Similarly for $S$. Now you can consider $R\times S$ as a $K$-algebra; a priori the structure could be different from the initial one: prove it is indeed the same.
Suppose $\operatorname{Mat}_n(D)$ is a $K$-algebra, where $D$ is a division ring. Then we have a ring homomorphism $K\to\operatorname{Mat}_n(D)$, with the image contained in the center. The center is formed by the scalar matrices with entries in the center of $D$, so it is isomorphic to the center of $D$. We use this isomorphism for endowing $D$ with the structure of $K$-algebra, which induces on $\operatorname{Mat}_n(D)$ the same structure of $K$-algebra we started with.
Best Answer
Here's my attempt at answering my own question based on the hints.
Let $x=\sum_{g\in S_3} g$. Then $yx=x$ and $xy=x$ for all $y\in S_3$ so $\mathbb{F}_5\cdot x=\{0,x,2x,3x,4x\}$ is a 2-sided ideal (elements of $\mathbb{F}_5[S_3]$ are sums of $y$'s so just act as a sum of identities).
But ideals of $\mathbb{F}_5[S_3]$ are $\mathbb{F}_5[S_3]$-submodules of $\mathbb{F}_5[S_3]$. Then $\mathbb{F}_5[S_3]$ is semisimple so each submodule is a direct summand, so there exists a term in the product isomorphic to $\mathbb{F}_5$.
So from the choices above we only have $\mathbb{F}_5\times \mathbb{F}_5 \times M_2(\mathbb{F}_5)$.