Short answer, yes.
In fact your "definition 2" gives $\mathbf a$ as the coefficients of the equation of a hyperplane as a nonzero vector, and this nonzero vector is a normal to that hyperplane as called for in "definition 3".
The algebra to prove this is very easy. Write down the equations, satisfied by $x$ and $y$ of the hyperplane respectively, and take the difference. We immediately have $\mathbf a \cdot (x-y) = 0$ as the constant $c$ cancels.
The argument goes the other way, to show that when $\mathbf b$ in "definition 3" is given, it also serves as a set of coefficients $\mathbf a$ in "definition 2". That is, suppose a hyperplane $p$ has a normal vector $b$ as described above, and fix some point $y$ in that hyperplane:
$$ \mathbf b \cdot (x-y) = 0 \; \text{ for any } x\in p$$
Expanding the above gives $\mathbf b \cdot x = \mathbf b \cdot y$, and now that $y$ is fixed we get an equation of the form in "definition 2" where constant $c = \mathbf b \cdot y$:
$$ \mathbf b \cdot x = c $$
Therefore any $\mathbf a$ that defines the hyperplane equation in "definition 2" will serve as the normal vector $\mathbf b$ defined in "definition 3", and any normal vector $\mathbf b$ that satisfies "definition 3" will serve as the vector of coefficients $\mathbf a$ in "definition 2". As a result these quantities have the same uniqueness property, namely that they are unique only up to taking a nonzero scalar multiple of a vector.
There seems to be a lesson here about rank or dimension of a subspace. The normal vector to a hyperplane is unique (up to a nonzero scalar multiple) in $\mathbb R^n$ because the hyperplane is an $n-1$ dimensional thing, and this leaves only one dimension for things to be perpendicular to this.
So the uniqueness you want to show depends on making use of the $n-1$ dimensional nature of the hyperplane. The easiest way to discuss this depends to a great extent on your background in linear algebra.
From the perspective of the one linear equation $\mathbf a \cdot x = c$, we see that the matrix form of this equation has a coefficient matrix of rank one (because $\mathbf a$ is nonzero). Therefore the solutions to this equation (via the principle of superposition) consist of a particular solution to the inhomogenous equation plus all solutions of the homogenous equation $\mathbf a \cdot x = 0$. Thus we have an $n-1$ dimensional "affine variety" of solutions.
But let's consider the uniqueness strictly from the perspective of the normal vector. To simplify things, the normal vectors to the hyperplane $\mathbf a \cdot x = c$ are exactly the same as the normal vectors to the subspace $\mathbf a \cdot x = 0$. This subspace has a basis of $n-1$ linearly independent vectors. There is a one dimensional subspace orthogonal to these, (orthogonal is another word for perpendicular or normal), and that subspace is generated by vector $\mathbf a$.
Taken together the basis for the subspace $\mathbf a \cdot x = 0$ and the normal vector $\mathbf a$ give us a basis for all of $\mathbb R^n$. The latter (because it is generated by one vector) is one dimensional, and the former is $n-1$ dimensional (so that the dimension of $\mathbb R^n$ is $n$ as we know).
This is a special case of the Rank-Nullity Theorem, which says the dimension of the row space of the coefficient matrix of the linear "system" $\mathbf a \cdot x = 0$ plus the dimension of the nullspace (the solutions/points of the hyperplane $\mathbf a \cdot x = 0$) is equal to the dimension $n$ of the domain $\mathbb R^n$.
Now the statement that all the nonzero normal vectors to the hyperplane are related by $\mathbf b = k\mathbf a$ for nonzero scalars $k$ is exactly the proposition that the subspace generated by one of these normal vectors is one dimensional. In other words Span($\mathbf b$) = Span($\mathbf a$) forces $\mathbf b$ to be a nonzero multiple of $\mathbf a$ and vice versa.
Best Answer
So to formalize discussion with Triatticus, proof is following:
If we have some hyperplane given by: $$H=\left\{ x\ \epsilon\ R^n : u^Tx=v \right\}$$
Then any two vectors $a,b$ which point to any separate points in this hyperplane are such that:
$$\begin{cases}u^Ta=v \\u^Tb=v \end{cases}$$
If we want to check whether hyperplane is orthogonal to normal vector $u$, then we have to find whether $u^T(a-b)=0$ for any a,b , because $(a-b)$ is a vector IN the hyperplane, whereas a and b are vectors which point in direction of points in the hyperplane.
If we solve above linear equations we have : $$u^T(a-b)=0$$ and this completes proof.