Why is there no vector component product operation? Considering two vectors
$$
\begin{align*}
\mathbf u &= \left\langle a,b,c \right\rangle \\
\mathbf v &= \left\langle d,e,f \right\rangle
\end{align*}
$$
we have the standard vector products given by
$$
\begin{align*}
\mathbf u \cdot \mathbf v &= ad+be+cf\\
\mathbf u \times \mathbf v &= \begin{vmatrix}
\hat\imath & \hat\jmath & \hat k\\
a&b&c\\d&e&f
\end{vmatrix}.
\end{align*}
$$
However, it seems that there could also easily be defined a vector component product, given by
$$
\begin{align*}
\mathbf u * \mathbf v &= \left\langle ad, be, cf \right\rangle.
\end{align*}
$$
Why is this not a standard operation? Is there no use for it in mathematics?
 A: You're more than welcome to define $\text{u} * \text{v}$ as you have.  That's what makes mathematics wonderful.
The question then becomes: "How will people use it?"  You're taking the component-wise product of the elements of two vectors, and creating another vecctor out of them.  What does it do?
I can see that if you take the vectors as being the diagonal elements of two square matrices, then your product vector is the components of the trace of the matrix product.
I see that $\text{u} * \text{v} = \text{v} * \text{u}$, so it's commutative.
I see that $\text{u} * \text{1} = \text{1}$, so it behaves well with the traditional multiplicative identity element.
I see that $\text{u} * \text{0} = \text{0}$, so it doesn't create any mathematical wormholes with the zero vector.
So, yeah:  you've created $*$, and it does some things.  What else does it do?
A: More generally, the Hadamard product of two $m\times n$ matrices fits your description in the sense that it is an entrywise product.
Thinking of $\mathbf{u}=(a,b,c)$ and $\mathbf{v}=(d,e,f)$ as $3\times 1$ matrices, their Hadamard product is the $3\times 1$ matrix $\mathbf{u}\circ\mathbf{v}=(ad,be,cf)$. 

Keep in mind that the definitions we choose are often a function of the theorems that we can prove from those definitions.
As far as what it is good for, note the structure present: it is associative, distributive, and even commutative(!). The downside is that composition of linear maps doesn't equate to this type of matrix multiplication and this is the thrust of "usual" matrix multiplication. 
