After a small discusion in the comments I decided to upload a complete answer to the problem, in case if anyone is intrested in the future for this counterexample. I will also point out my misinterpretation which was in the definition of simplicity.
The problem: Find an example of a simple ring which is not semisimple.
Let me clarify the above definitions first. We say that the ring $R$ is simple if $R$ does not have any other two-sided ideals other than $0$ and itself. We say that $R$ is (left) semisimple if for every (left) ideal $I$ there exists a (left) ideal $J$ with $R=I\oplus J$. I had confused the definition of simplicity with the following weaker condition which is known as left simplicity. We say that the ring $R$ is left semisimple if $R$ does not contain any other left ideals other than $0$ and itself.
Solution to the problem: Let $\mathbb{F}$ be a field and $V$ a vector space over $\mathbb{F}$ with a countably infinite basis. Let $R=End_\mathbb{F} V$ be the ring of all linear operators from $V$ to $V$. Furthermore, let $I$ be the two-sided ideal of $R$ which constitutes of all linear operators $x:V\to V$ with finite dimensional image. I.e.
$$I=\biggl\{x\in R:\,\text{dim im}x<\infty\biggr\}.$$
We claim that the quotient space $R/I$ is our desired example. The result will follow immediately from the following claims:
Claim 1: For every $f\in R\setminus I$ there exits $x,y\in R$ with $xfy=I_V$, where $I_V$ denotes the identity operator on $V$.
Proof of Claim 1: Let $f\in R\setminus I$. Then $\text{dim im }f=\infty$. Let $(e_n)_{n=1}^{\infty}$ be a countably infinite basis of $\text{im} f$. Then, there exists $u_n$ with $f(u_n)=e_n$. We extend the basis (possibly by using Zorn's lemma) of $\text{im} f$ to a basis of the whole space $V$. Therefore, we can adjust some elements $(v_j)_{j\in J}$ indexed by some set $J$ such that the collection
$$\{e_n:\,n\in \mathbb{N}\}\cup \{v_j:\,j\in J\}$$
is a basis of $V$. By our hypothesis that $V$ has a countable basis $J$ would be at most countably infinite. We consider two cases:
Case 1: J is countably infinite. In this case we write $(v_j)_{j\in J}$ as a sequence $(v_n)_{n\in \mathbb{N}}$. We rewrite the basis in a sequence $(e_n')_{n\geq 1}$ with $e_{2n}'=e_n$ and $e_{2n-1}'=v_{n}$ for every $n\geq 1$. To define the operators $x,y$ we only need to determined their values on the elements of the basis. We define $x,y$ by the following relations:
$$e_{2n}'\overset{y}{\longrightarrow}u_{2n}\overset{f}{\longrightarrow}e_{2n}=e_{4n}'\overset{x}{\longrightarrow}e_{2n}'$$
$$e_{2n-1}'\overset{y}{\longrightarrow}v_{2n-1}\overset{f}{\longrightarrow}e_{2n-1}=e_{4n-2}'\overset{x}{\longrightarrow}e_{2n-1}'.$$
Note that $y$ has been defined in all elements of the basis, hence it can be extended uniquely to a linear operator $y:V\to V$. We define $x$ on the rest of the basis elements by giving an arbitrary value and we extend $x$ also. Now the extended operators, which we denote again by $x,y$ satisfy $(x\circ f\circ y)(e_n')=e_n'$ for every $n$. Therefore, for every $v\in V$ we have $(x\circ f\circ y)(v)=v$. In other words, $xfy=I_V$.
Case 2: J is finite. We treat this case in similar fashion as the previous one, I will omit this step to make the answer a little more compacted.
Claim 2: $R/I$ is a simple ring.
Proof of Claim 2: Let $0\neq \overline{J}\subseteq R/I$ be a two-sided ideal of $R/I$. Then there exists a two-sided ideal $I\subsetneq J\subseteq R$ with $\overline{J}=J/I$. Now, let $0\neq f+I\in \overline{J}$. Then, $f\in J\setminus I$, thereby $\text{dim im} f=\infty$. By the preceding claim there exists $x,y\in R$ with $xfy=I_V$. Since $J$ is a two-sided ideal, it follows that $xfy\in J$, consequently $I_V+I\in \overline{J}$ and from this we obtain that $\overline{J}=R/I$.
Claim 3: $R/I$ is not left Notherian neither left Artinian.
Proof of Claim 3: For every subspace $U\subseteq V$ let $J_U$ be the left ideal of $V$ given by
$$J_U=\{x\in R:\, x(U)=0\}.$$
It is easily seen, that for every two subspaces $W\subseteq U\subseteq V$ we have the following inclusions $J_U\subseteq J_W$. Futhermore, if $\text{dim}_{\mathbb{F}}U/W=\infty$ then $I+J_U\subsetneq I+J_W$. Indeed, first we begin with a basis $(e_i)_{i\in I}$ of $W$ then we extend this basis to a basis $(e_i)_{i\in I}\cup (v_j)_{j\in J}$ of $W$. By the fact that $\text{dim}_{\mathbb{F}}U/W=\infty$ it follows that $J$ is countably infite. At the last step, we extend the basis $(e_i)_{i\in I}\cup (v_j)_{j\in J}$ to a basis $(e_i)_{i\in I}\cup (v_j)_{j\in J}\cup (u_k)_{k\in K}$ of the whole space $V$. We define a linear operator $f:V\to V$ by the relations $f(e_i)=0,\, f(v_j)=v_j$ and $f(u_k)=u_k$. Then, obviously $f\in J_W$ and thereby $f\in I+J_W$. We claim that $f\notin I+J_U$. If not, then $f=x+y$ for some $x\in I$ and $y\in J_U$. Then, on one hand $\text{dim im}(f-y)<\infty$ and on the other hand $x(v_j)=f(v_j)-y(v_j)=f(v_j)=v_j$ and thereby $v_j \in \text{im }x$. But this a contradiction since $\text{dim im} x<\infty$ and $J$ is countably infinite.
Now can prove that $R/I$ is not left Notherian for example. We fix a countable basis $(e_n)_{n=1}^{\infty}$ of $V$. Let a descending sequence
$$J_1\supseteq J_2\supseteq ...\supseteq J_n\supseteq J_{n+1}\supseteq ...$$
of infinite subsets of $\mathbb{N}$ with the property that $J_n\setminus J_{n+1}$ is infinite. For every $n$ we define the subspace $W_n=\text{span}\{e_m:\,m\in J_n\}$ of $V$. Then since $(J_n)$ is descending it follows that $W_n\supseteq W_{n+1}$. The fact that $J_n\setminus J_{n+1}$ is infinite implies that $\text{dim}_{\mathbb{F}}W_n/W_{n+1}=\infty$. Consequently, $I+J_{W_n}\subsetneq I+J_{W_{n+1}}$. Therefore, if $\overline{J_n}=(I+W_n)/I$ then every $\overline{J_n}$ is a left ideal of $R/I$ and
$$\overline{J_1}\subsetneq \overline{J_2}\subsetneq...\subsetneq \overline{J_n}\subsetneq \overline{J_{n+1}}\subsetneq...$$
and thereby $R/I$ cannot be left Notherian. In similar fashion we show that $R/I$ is not left Artinian too.
Any comments or corrections are appreciated!
Best Answer
The one in Wikipedia is the standard definition of a simple ring, but from what you describe, Lang's definition amounts to what is more commonly called a simple Artinian ring.
For a simple ring (by which I mean the definition appearing in Wikipedia) the conditions of being right Artinian or left Artinian are equivalent. As a special case of the Artin-Wedderburn theorem, such a ring is just $M_n(D)$ for some division ring $D$.
Perhaps it was Lang's intention to define a simple ring such that all of his simple rings are semisimple too. This is a point of terminological confusion sometimes, because the more common definition allows simple rings which aren't Artinian, hence aren't semisimple.
I don't know if anyone has tried it, but maybe it would make sense to call finite products of simple rings "semisimple" and to give "semisimple Artinian rings" another name, say "Wedderburn ring" or something like that. Isaacs called them Wedderburn rings in his Algebra book, actually, but I don't think it caught anywhere else.