I am trying to show that if a ring $R \cong M_n (D)$ for some division ring $D$ (meaning isomorphic as rings) then $R=S \oplus … \oplus S$ ($n$ times).
Firstly I can see that $M_n(D) = C_1 + …+ C_n $ where $C_i $ is the ith column and there are just zeroes in all the other columns. I know that each column is isomorphic to $D^n $ which is simple as a left $M_n (D)$-module.
What I want to show that somehow using the isomorphism from $M_n(D)$ to $R$ we can use this to show the result but I’m a bit confused.
Yes, and in fact the sum is direct.
Perhaps what you are missing is that $M_n(D)$ only has one simple module (up to isomorphism)? There are a couple ways to do this but the easiest is probably to see why there's only one Wedderburn component for this ring (because it is simple.)
The core question here is "If $\phi:R\to S$ is a ring isomorphism, and $f:M\to N$ is an isomorphism of $R$ modules, then in what sense are $M$ and $N$ isomorphic as $S$ modules?"
The only natural $S$ module structure available to $M$ and $N$ is that of transfer (via the ring isomorphism), where the action is defined by $\phi(r)\cdot x=rx$ (where $x$ comes from $M$, or $N$, whichever you have settled on.) You can check to see that all the $S$-module axioms are satisfied if you start with an $R$ module. I use the dot here to help distinguish the $R$ action on $M$ from the $S$ action on $M$.
You can compatibly lift $f$ to be an $S$ module homomorphism with this action. $f$ is already a group isomorphism of $M$ and $N$. Does it respect the transferred action?
$\phi(r)\cdot f(m):=rf(m)=f(rm)=f(\phi(r)\cdot m)$
The first and last equalities are the definition of the transferred action, the middle one is $R$ linearity of $f$. So $f$ is $S$ linear too.
Now you should be comfortable concluding that if $M_n(D)$ is isomorphic to a direct sum of $n$ mutually isomorphic simple left ideals, and $R$ is ring isomorphic to $M_n(D)$, then $R$ is also a direct sum of $n$ mutually isomorphic simple left ideals.