In different categories (such as the category of rings or the category of modules) different substructures are important. Ideals are the natural substructures of rings, and submodules are the natural substructures of modules. These substructures are special because they are suitable for forming quotient objects in their categories (that is why I am not mentioning 'subrings,' for example.)
"Simple" has come to mean "having no nontrivial substructure," (or as suggested in the comments "admitting no nontrivial quotient") so in the case of rings this means it has no nontrivial ideals.
If you want to talk about right and left ideals, then you must realize that we are switching gears and thinking of the ring as a module over itself and no longer like a ring. Notice how the definition you gave for semisimplicity of a ring is entirely in terms of its semisimplicity as a module.
Anyhow, this gives you the answer to your question. The definition of a "simple ring" refers only to ring ideals, and that is not directly connected to its identity as a module.
How to construct such a ring?
The Artin-Wedderburn theorem completely classifies all semisimple rings. They are exactly finite direct products of matrix rings over division rings. This gives you every possible example.
You can make simple rings out of any ring by taking the quotient by a maximal ideal. To get an example where the result isn't semisimple, you'll just have to be careful to make sure the quotient isn't Artinian. One such example is to take the ring of linear transformations of an infinite dimensional vector space and take the quotient by a maximal ideal. This is a non-semisimple simple ring.
There is even a simple ring without nonzero zero divisors. One such example is that of Weyl algebras. Semisimple rings usually have nonzero zero divisors: the only semisimple rings without nonzero zero divisors are division rings.
When will the result hold?
A simple ring will be semisimple iff any if the following conditions hold:
it has a minimal left ideal (or minimal right ideal)
It is right artinian (or left artinian)
The proof is correct, although it would be preferable to explain why the zero ideal is a finite intersection of maximal right ideals a little more carefully. It is true that in an artinian poset of one sided ideals, a collection intersecting to zero must have a finite subset intersecting to zero, but it is not really done justice by saying “the set had a minimal element which is obviously (read “necessarily”) zero.”
My first thought was to take $R→⨁_M R/M$ [...] with $M$
running over all maximal left ideals, but I realized that I couldn't guarantee that this was a direct sum a priori.
Well, you can make that sum and it will be direct, nothing prevents you from doing so. But yes, the map you want to do (sending $r$ to its image in the quotient) doesn’t work. You would necessarily be working in $\prod_M R/M$ instead. Reducing it to a finite product makes it equivalent to a direct sum, and that was a very important reduction.
Is there a similar argument to prove that there are only finitely many maximal ideals in the noncommutative case?
Yes, the same argument (with noncommutative prime ideals) works. If a prime ideal contains a finite intersection of prime ideals, it necessarily contains their product, whence it contains at least one of the terms the product.
Best Answer
I'll make a try to write correctly an answer I came across:
Let $R$ Artinian with $rad(R)=0$ and $R$ not ss (semisimple). I choose $$I_1\subset R$$ minimal left ideal. It is $$I_1\not\subset 0=rad(R)$$ so $\exists$ $M_1$ maximal left ideal of $R$ such that $$I_1\not\subset M_1.$$ Therefore $I_1\cap M_1 \subset I_1 \Rightarrow I_1\cap M_1=0 $ and $M_1\subset I_1+M_1$ so $I_1\oplus M_1=R$. Since $M_1\not=0$ there exists a minimal left ideal $I_2\subset M_1$. Similarly I can find a maximal left ideal $N_2\subset R$ st $R=I_2\oplus N_2$. Since $I_2\subset M_1$ and $R=I_2\oplus N_2$ it is $M_1=I_2\oplus (N_2\cap M_1)$. So I write $M_2=N_2\cap M_1$ and I have $M_1=I_2\oplus M_2$.
So far we have $R=I_1\oplus I_2\oplus M_2$ and $M_2\not=0$. We repeat the same process. The sequence of left ideals $$R\supset M_1\supset M_2\supset ...$$ is strictly decreasing which is a contradiction.
Is this correct? Is there a shorter way to prove this?