$A+B=AB$ does it follows that $AB=BA$? If $A$, $B$ are two normal operators such that: $A+B=AB$ does it follow that $AB=BA$?

First, note that if $C$ is a normal operator which has a left inverse $D$ (so $DC=I$), then $C$ is invertible (and thus $D$ is its inverse and $CD=I$ as well).  This follows from the spectral theorem: you can identify your Hilbert space with $L^2(X)$ for some semifinite measure space $X$ and $C$ with multiplication by $f$ for some $f\in L^\infty(X)$.  If $0$ is in the essential range of $f$, then this means that for any $\epsilon>0$ there exists $g\in L^2(X)$ such that $\|fg\|<\epsilon\|g\|$ (choose $g$ to be supported on the set where $|f|<\epsilon$).  Since $DC=I$, we must have $D(fg)=g$ so $\|D\|\geq\|g\|/\|fg\|>1/\epsilon$.  Since $\epsilon$ is arbitrary, this is a contradiction.  Thus $0$ is not in the essential range of $f$, so $1/f\in L^\infty(X)$, and multiplication by $1/f$ is an inverse for $C$.
(Probably there is a more elementary argument that avoids spectral theory, but this is all I can come up with at the moment.)
Now note that $A+B=AB$ implies $(I-A)(I-B)=I-A-B+AB=I$.  Since $I-B$ is normal, by the result above this implies $(I-B)(I-A)=I$ as well.  That is, $I-A-B+BA=I$, or $BA=A+B=AB$.
