Left Multiplication Ring Homomorphism 
Assume we have a non-commutative unital ring $R$ and an element $r$ not in the center. Define a map $$\phi_r:R\rightarrow R$$ $$x\mapsto rx$$ Can this ever be a ring homomorphism?  

If it can be then $r$ has to be idempotent, and the kernel (the set of $r$'s zero divisors with $0$) must be an ideal.  I'm pretty sure that this can't be a homomorphism, but I can't prove it.  Although I can't promise that I just haven't been clever enough to think of an example where this works.  Some simple proof showing that $r$ has to be in the center is what I've striven for, but alas...
Edit:  Suppose I should add that I'm adopting the definition of a ring homomorphism that isn't necessarily (multiplicative) identity invariant.
 A: Let $R$ be the ring of $2\times 2$ upper triangular matrices with coefficients in some field, and let $r=\begin{pmatrix}0&0\\0&1\end{pmatrix}$.
A: It is certainly possible to concoct an artificial example where $x\mapsto rx$ is a homomorphism but $r$ does not commute with everything.
The elements shall be non-commutative polynomials in two variables, say $x$ and $y$, such that no term of the polynomial contains more than one occurrence of $x$. Addition is ordinary polynomial addition. For multiplication, multiply the polynomials as usual and then delete from each term the second occurrence of $x$, if there is one.
For example:
$\begin{array}{rcl}(2y^2xy + x + 2y)(xy+yx)&=&2y^2xyxy + x^2y + 2yxy + 2y^2xy^2x + xyx + 2y^2x \\&=& 2y^2xy^2 + xy + 2yxy + 2y^2xy^2 + xy + 2y^2x \\&=& 4y^2xy^2 + 2xy + 2yxy + 2y^2x\end{array}$
In other words, this is the free ring on $x$ and $y$ quotiented by the relations $xab=xaxb$ for all $a$ and $b$.
Clearly $x$ does not commute with $y$, so is not in the center, and yet $xab=(xa)(xb)$ for all $a$ and $b$.
A: Let's assume that $x\mapsto ex$ is a ring homomorphism. Then of course the image $eR$ is a ring with identity $e$, and so (as you observed) it will be idempotent and central in the ring $eR$. Those are certainly necessary conditions, so it's natural to ask if they're sufficient also.
Suppose that $e^2=e$ and $e$ commutes with everything in $eR$. Then $x\mapsto ex$ is obviously an additive homomorphism from $R$ to $eR$, and additionally $(ex)(ey)=(exe)y=(eex)y=exy$, so it's a multiplicative homomorphism as well. Thus $eR$ is a ring with identity $e$, and the multiplication map is a ring homomorphism.
So we have the answer:

The map $x\mapsto ex$ from $R\to R$ using the element $e\in R$ is a ring homomorphism iff $e$ is an idempotent of $R$ commuting with each element of $eR$.

In particular, any idempotent that's in the center of $R$ will create the ring $eRe=eR$ inside of $R$. Mariano's example shows us a noncentral idempotent satisfying the condition highlighted above. If you look at the same idempotent in the full ring of $2\times 2 $ matrices, you'll find it fails to commute with $eR$. (Just check its product with $\begin{bmatrix}0&0\\1&1\end{bmatrix}$ on both sides, for example.)
