In Algebra by Serge Lang (XVII, Proposition 1.2, p. 643-644) it is shown that for a simple $R$-module $E$ and $n, m \geq 0$ with $E^{\oplus n} \cong E^{\oplus m}$ it follows that $n = m$. Lang argues as follows:
[…] $\operatorname{End}_R(E^{\oplus n})$ is isomorphic to the $n \times n$ matrix ring over the division ring $\operatorname{End}_R(E) = K$. Furthermore, this isomorphism is verified at once to be an isomorphism as $K$-vector spaces. The dimension of the space of $n \times n$ matrices over $K$ is $n^2$. This proves that the multiplicity $n$ is uniquely determined […].
I don’t see why this isomorphism (of rings) should be $K$-linear:
We need to show that the resulting isomorphism $\operatorname{Mat}_n(K) \to \operatorname{Mat}_m(K)$ maps every scalar matrix $\lambda I_n$ to the corresponding scalar matrix $\lambda I_m$. For this we will need to use that this isomorphism has a special form, because this does not hold for every isomorphism (complex conjugation $\operatorname{Mat}_n(\mathbb{C}) \to \operatorname{Mat}_n(\mathbb{C})$ gives a counterexample):
If $f \colon E^{\oplus n} \to E^{\oplus m}$ is an isomorphism of $R$-modules then we can represent $f$ by a matrix $S \in \operatorname{Mat}(m \times n, K)$ and $f^{-1}$ by a matrix $T \in \operatorname{Mat}(n \times m, K)$, for which we then have that $ST = I_m$ and $TS = I_n$. The isomorphism $\operatorname{Mat}_n(K) \to \operatorname{Mat}_m(K)$ in question is then given by $$ A \mapsto SAT = S A S^{-1}. $$ I don’t see why this should map $\lambda I_n$ to $\lambda I_m$, because $\lambda$ does not have to commute with $S$.
If I’m not mistaken then one can use the invertibility of the matrix $S$ to conclude that $n = m$ (because it follows that $K^n \cong K^m$ as right $K$-vector spaces), which proves the proposition itself. But even then I don’t see why the isomorphism $\operatorname{Mat}_n(K) \to \operatorname{Mat}_m(K) = \operatorname{Mat}_n(K)$ should be $K$-linear.
Any help is welcome.
I may have found an answer. I do not believe the Lang's argument is quite right. Instead, I have used the following line of proof:
Let $E$ be a left simple $R$-module. By the Schur's lemma, the ring $K := End_{R}(E)$ is a division ring.
Consider an $R$-linear map $\varphi: E^{n} \rightarrow E^{m}$. One can assign to it a matrix $\mathbf{M}(\varphi) \in K^{m,n}$, where $K^{m,n}$ is the space of $m \times n$ matrices with entries in the ring $K$. Indeed, due to $R$-linearity, one can write the action of $\varphi$ on $(x_{1},\dots,x_{n}) \in E^{n}$ as
$\varphi(x_{1},\dots,x_{n}) = (\varphi_{11}(x_{1}) + \dots + \varphi_{1n}(x_{n}), \dots, \varphi_{m1}(x_{1}) + \dots + \varphi_{mn}(x_{n}))$
for a unique collection $\{ \varphi_{ij} \} \subseteq K$. This defines the desired matrix $\mathbf{M}(\varphi) \in K^{m,n}$. In fact, observe that if $\psi: E^{m} \rightarrow E^{s}$ is another $R$-linear map, one has $\mathbf{M}(\psi \circ \varphi) = \mathbf{M}(\psi) \mathbf{M}(\varphi)$, where on the right hand side, there is a matrix multiplication of a matrix from $K^{s,m}$ with the matrix from $K^{m,n}$.
On the other hand, any matrix from $\mathbf{A} \in K^{m,n}$ corresponds to a unique $K$-linear map $f_{\mathbf{A}}: K^{n} \rightarrow K^{m}$ of right $K$-modules $K^{n}$ and $K^{m}$ (with obvious right action of $K$). Indeed, if $(e_{1},\dots,e_{n})$ is the standard basis of $K^{n}$, then $f_{\mathbf{A}}( e_{i} \cdot k^{i}) := f_{j} \cdot (\mathbf{A}_{ji} \cdot k^{i})$, where $(f_{1},\dots,f_{m})$ is the standard basis of $K^{m}$. This assignment respects the matrix multiplication, that is if $\mathbf{B} \in K^{s,m}$, then $f_{\mathbf{B} \mathbf{A}} = f_{\mathbf{B}} \circ f_{\mathbf{A}}$. Note that this is the reason why we have to view $K^{n}$ and $K^{m}$ as right $K$-modules (since $K$ is in general not a commutative ring).
A combination of 2. and 3. shows that to any $R$-linear map $\varphi: E^{n} \rightarrow E^{m}$, one can assign a matrix $\mathbf{M}(\varphi) \in K^{m,n}$ and to it a unique $K$-linear map $\hat{\varphi} := f_{\mathbf{M}(\varphi)}: K^{n} \rightarrow K^{m}$. Moreover, if $\psi: E^{m} \rightarrow E^{s}$ is another $R$-linear map, we see that $\hat{\psi \circ \varphi} = \hat{\psi} \circ \hat{\varphi}$.
Finally, if $\varphi: E^{n} \rightarrow E^{m}$ is an $R$-linear isomorphism, the part 4. shows that $\hat{\varphi}: K^{n} \rightarrow K^{m}$ is a $K$-linear isomorphism. But this is only possible if $n = m$.