Let $D$ be a division ring and consider $D$ as a right module over itself. Now consider the endomorphism ring of this module. If $s$ is some endomorphism, then $s(1) = a = a1$ for some $a\in D$ so the map $x \mapsto s(x) - ax$ is a morphism with non-trivial kernel, and hence $0$. This means that $s$ is simply multiplication by $a$ from the left, and we get that the endomorphism ring of $D$ as a module over itself is isomorphic to $D$.
The answer is that "yes, $M$ is isomorphic to $I_j$ (for any $j$!), but it is not necessarily 'induced' by a projection $R\to M$."
Given any simple left module $S$, a projection $\phi: R\to S$ turns into an isomorphism $R\ker\phi\cong S$. Where $\ker\phi$ is a maximal left ideal of $R$. Since $R$ is semisimple, $R\cong \ker\phi\oplus N$ where $N$ is a left ideal of $R$. So $S\cong R/\ker\phi=(N\oplus\ker\phi)/\ker\phi\cong N$.
This shows that every simple left module appears as a minimal left ideal of $R$. But $N$ need not be one of the $I_j$. For example, $N$ could be the set of matrices which have an arbitrary first column, the second column identical with the first, and the rest of the columns zero. There are lots of other simple left ideals of $R$ other than the $I_j$.
To prove that all simple left $R$ modules are mutually isomorphic, we can use this
Lemma: For a fixed minimal left ideal $N<R$, the sum $I_N=\sum\{N'<R\mid N\cong N'\}$ is an ideal of $R$ ($N'$ denotes a left ideal, of course.) Furthermore, if $L$ is another minimal left ideal which is nonisomorphic to $N$, then $I_L\cap I_N=\{0\}$.
Since $R$ is simple, only one such $I_N$ can exist, and that means there is only one isotype of simple left $R$ module. That is why all of the $I_j$ in your original post are mutually isomorphic, and indeed every other simple left $R$ module you find is going to be isomorphic to these also.
Best Answer
It can very easily verified that every non-zero endomorphism $\varphi$ over simple $R$-module $M$ is bijective.
Because these two holds...
$$0\le\ker\varphi\ne M, 0\ne\mathrm{im}\varphi\le M$$ and $R$-sumbodule of $M$ is only $0, M$.
But it is not a field. If $R$ is non-commutative, then there exists $a,b\in R$ such that $ab\ne ba$, and $L_a, L_b:M\to M$ such that $L_a(m)=am, L_b(m)=bm$ are not commutative.
When $R$ is commutative, then there exists maximal ideal $m$ such that $M\cong\frac Rm$ by this. And $\frac Rm$ is field!
Every endomorphism over field is $L(x)=kx$ form. And left is easy to verify.