Actually, your version is correct and Wikipedia's is wrong! (Except that it has a tiny disclaimer at the bottom.)
If $\mathrm{P}=\mathrm{NP}$, Wikipedia claims that every problem in $\mathrm{P}$ is $\mathrm{NP}$-complete. However, this is not true: in fact, every problem in $\mathrm{P}$ would be $\mathrm{NP}$-complete, except for the trivial languages $\emptyset$ and $\Sigma^*$.
You can't many-one reduce any nonempty language $L$ to $\emptyset$, because a many-one reduction must map "yes" instances of $L$ to "yes" instances of $\emptyset$, but $\emptyset$ has no "yes" instances. Similarly, you can't reduce to $\Sigma^*$ because there's nothing to map the "no" instances to. However, if $\mathrm{P}=\mathrm{NP}$, then every other language in $\mathrm{P}$ is $\mathrm{NP}$-complete, since you can solve the language in the reduction.
So, just to make it explicit:
- your diagram is correct;
- Wikipedia's isn't (unless you read the tiny disclaimer);
- the area you've labelled "$\mathrm{P}$, $\mathrm{NP}$" contains the two languages $\emptyset$ and $\Sigma^*$, and nothing else;
- the area you've labelled "$\mathrm{P}$, $\mathrm{NP}$-complete" contains every other language in $\mathrm{P}$ and nothing else.