is every element in a C* algebra a product of normal elements?

The shift in $\ell^2$ cannot be a product of normal operators, since its Fredholm index is nonzero, and a normal operator cannot have nonzero Fredholm index.

Since the question was asked wrt C$*$-algebras, I guess there is room for a general remark. Suppose that $xy = 1$ where $x$ is a product of normal elements, say $v_1v_2\cdots v_n$. Then $v_1$ is normal and has a right inverse; therefore it is (two-sided) invertible. Thus $v_2\dots v_n y $ is invertible, so $v_2$ has a right inverse, and thus is invertible. By induction, we obtain that all the $v_i$ are two-sided invertible, so that $x$ is too.

The upshot of this argument is that a product of normal elements that is one-sided invertible, is actually invertible. In particular, the right shift is not a product of normal elements---although of course, the index argument above is decisive in this case.

If every element of a C$*$-algebra were a product of normals, then it would be directly finite (all one-sided inverses are two-sided). Whether this is sufficient is unclear (as is the question as to whether the product of normals property extends to matrix rings over the original). But if the C*-algebra has one in the stable range, then every element is an invertible times a positive, so is a product of (three) normals.

And for von Neumann algebras, if of finite type, then every element is a product of two normals, as the isometry in the polar decomposition argument (below) can be modified to be a unitary in that case. If not of finite type, the property fails.

Recall the polar decomposition: Let $T$ be any operator on a Hilbert space, let $|T|=\sqrt{T^*T}$. Then there exists an isometry $U$ from the image of $|T|$ to the image of $T$ such that $$ T=U|T|. $$ Now, whenever $U$ may be extended to a unitary on $H$, we are done. This is equivalent to saying that the image of $|T|$ and the image of $T$ have the same codimension. A non-zero Fredholm-index is an obstacle to that.