$$Ax=b$$
$A$ is a real square symmetric matrix with all non-zero entries, $x$ and $b$ are vectors.
The hadamard inverse of $A$ is defined
$$A^{\circ-1} \circ A = E$$
where $E$ is a matrix of ones.
So, multiplying the LHS of the first equation by $A^{\circ-1}$ gives $Ex$. That much is pretty clear. But what happens when you multiply the RHS by $A^{\circ-1}$? The operation $A^{\circ-1}\circ b$ doesn't seem well defined since there is a dimension mismatch.
However, considering that $b=Ib$, where $I$ is the identity matrix, perhaps what you're supposed to do is
$$A^{\circ-1} \circ I b$$
But then you lose all the off diagonal elements of $A^{\circ-1}$ and you get
$$Ex=D_{A^{\circ-1}}b$$
(where $D_{A^{\circ-1}}$ is a diagonal matrix with the diagonal elements of $A^{\circ-1}$ and zeros elsewhere)
Which means
$$\sum_i x_i = A^{\circ-1}_{kk}b_k \:\:\: ; \:\: \forall k$$
So the elements of $b$ divided by the corresponding diagonal elements in $A$ have to be constant? That can't be right.
No idea what you're trying to do, but here's an idea for using a Hadamard product to solve the equation.
A Hadamard inverse (let's call it $q$) exists for the vector $b$, assuming that all elements are non-zero. So we have the identity $$b\circ q = 1$$ Multiply both sides of the linear equation by this vector $$\eqalign{ q\circ(Ax) &= q\circ b = 1 \cr QAx &= 1 \cr x &= (QA)^{-1}\,1 \cr }$$ where $$Q = {\rm Diag}(q)$$