If
$\nabla \times \vec E = 0, \tag 1$
we know from a standard result that
$\vec E = \nabla \phi \tag 2$
for some scalar function $\phi$.
If the divergence of $\vec E$ is also a known function $\rho$,
$\nabla \cdot \vec E = \rho, \tag 3$
then combining (2) and (3) we obtain
$\nabla^2 \phi = \nabla \cdot \nabla \phi = \nabla \cdot \vec E = \rho, \tag 4$
which can in principle be solved for $\phi$ (assuming we admit appropiate boundary conditions on $\phi$; $\phi(x) \to 0$ sufficiently fast as $\vert x \vert \to \infty$ is often used; Jackson's book covers this, mos' likely); thus, we may discover $\vec E$ from (2).
We observe that such a solution is not unique; indeed, let $\psi$ be any harmonic function,
$\nabla \cdot \nabla \psi = \nabla^2 \psi = 0; \tag 5$
then
$\nabla^2(\phi + \psi) = \nabla^2 \phi + \nabla^2 \psi = \rho + 0 = \rho, \tag 6$
and if
$\vec E = \nabla (\phi + \psi), \tag 7$
we still obtain
$\nabla \times \vec E = \nabla \times \nabla (\phi + \psi) = \nabla \times \nabla \phi + \nabla \times \nabla \times \psi = 0, \tag 8$
since the curl of any gradient vanishes. Also,
$\nabla \cdot \vec E = \nabla \cdot (\nabla \phi + \nabla \psi) = \nabla^2 \phi + \nabla ^2 \psi = \nabla^2 \phi = \rho; \tag 9$
these last two equations show that we may transform any solution according to
$\phi \to \phi + \psi, \tag 9$
$\vec E = \nabla \phi \to \nabla \phi + \nabla \psi, \tag{10}$
$\psi$ as in (5), and preserve the divergence and curl of $\vec E$; so any solution is not unique; uniqueness may be attained by specifying appropriate boundary conditions on $\phi$ and $\psi$ which can then become unambiguously determined.
The above discussion addresses the relatively simple case (1), (3); we can, in fact, also address the significant generalization
$\nabla \times \vec E = \vec F, \; \nabla \cdot \vec E = \rho, \tag{11}$
where $\vec F$ is a pre-specified vector field; the situation is more complicated since we may no longer assume $\vec E$ is a gradient as in (1)-(2). In this case we
instead invoke the vector calculus identity
$\nabla \times (\nabla \times \vec A) = \nabla (\nabla \cdot \vec A) - \nabla^2 \vec A, \tag{12}$
where the Laplacian operator $\nabla^2$ occurring on the right-hand side is understood to act component-wise on $\vec A$; thus we have
$\nabla \times \vec E = \vec F \tag{13}$
leading to
$\nabla \times (\nabla \times \vec E) = \nabla \times \vec F, \tag{14}$
which we trasform according to (12):
$\nabla (\nabla \cdot \vec E) - \nabla^2(\vec E) = \nabla \times \vec F; \tag{15}$
by virtue of (3) we write this as
$\nabla \rho - \nabla^2 \vec E = \nabla \times \vec F, \tag{16}$
whence we find
$\nabla^2 \vec E = \nabla \rho - \nabla \times \vec F, \tag{17}$
which we may in principle solve component-wise for $\vec E$; for example,
$\nabla^2 E_x = \dfrac{\partial \rho}{\partial x} - \left ( \dfrac{\partial F_z}{\partial y} - \dfrac{\partial F_y}{\partial z} \right) = \dfrac{\partial \rho}{\partial x} -
\nabla^2 E_x = \dfrac{\partial \rho}{\partial x} - \dfrac{\partial F_z}{\partial y} + \dfrac{\partial F_y}{\partial z}; \tag{18}$
of course, we need boundary conditions and perhaps certain restrictions on $\rho$ and $\vec F$; but I'll leave it to folks like David Jackson to explain such matters.
Apparently solutions to (11), (18) are still not unique; it is still true that adjusting $\vec E$ by the gradient of a harmonic function $\psi$ gives rise to another solution; of course, we may determine such $\psi$ uniquely via appropriate boundary conditions, and then a certain uniqueness is attained for $\vec E$.
Best Answer
First look at what $\hat{r}=(\hat{x},\hat{y})$ means.$$\hat{r}=(\hat{x},\hat{y})=\hat{x}\otimes \hat{x}+\hat{y}\otimes \hat{y}$$ Where $\otimes$ denotes vector scalar multiplication. Notice that by definition the vector scalar multiplication is a functions that takes in a scalar and a vector and outputs a vector, which is not the case here. So this definition as is doesn't make any sense because you have some undefined vector vector multiplication. You may define itsuch that this gives you a certain vector (and there are vector vector multiplications defined), and only then would it make sense to talk about $\hat{r}$; otherwise, you are in of course no position to talk about dot products.
Just for example we could try defining it like this: $$u\otimes v=A\hat{x}+B\hat{y}+C\hat{z}$$ Where $A,B,C$ are scalars. All that would be left would be to define what those scalars are. One such candidate is the cross product which would have $A,B,C$ in terms of the components of $u$ and $v$. If such a definition is given, then your $\hat{r}$ is the zero vector, so this probably wouldn't be very helfpul. Let's try and define it in a way that agrees with your integral. That is, $$\hat{x}\otimes \hat{x}=\hat{x} \hspace{1mm} \text{and} \hspace{1mm}\hat{y}\otimes \hat{y}=\hat{y}$$ So, if we adopt usual multiplication properties like associativity and distributivity $$u \otimes v=\left(\sum_i u_i \hat{x_i}\right)\otimes \left(\sum_i v_i \hat{x_i}\right)=\\ \sum_i \sum_j u_i\,v_j \hat{x_i} \otimes \hat{x_j} $$ Notice that when $i=j$ you have $\sum_i v_i\,u_i \hat{x_i} $. Now for $i \neq j$ you have to define something. We could just set it to zero. Then, your $\hat{r}$ becomes $$\hat{r}=\hat{x}+\hat{y}$$ and your integral becomes $\int dE_x+dE_y$.
Another thing to have in mind is that dot products are bilinear applications from pairs of vectors to scalars. As such, the claim that $(\hat{x},\hat{y}) \cdot (dE_x,d_Ey)=\hat{x}dE_x+\hat{y}dE_y$ doesn't make sense; granted that, if you want to, you can define another operation with this property, but it will no longer be a dot product.
Anyway the moral of the story is, you can do whatever you want as long as you define it well.Nonetheless, I would advise you to be careful when inventing definitions so that notations make sense. There have been times when it has yielded powerful math, but most of the times it just induces you in error.