Here is an algebraic approach to adjoint operators. Let us strip away the existence of an inner product and instead take two vector spaces $V$ and $W$. Furthermore, let $V^*$ and $W^*$ be the linear duals of $V$ and $W$, that is, the collection of linear maps $V\to k$ and $W\to k$, where $k$ is the base field. If you're working over $\mathbb R$ or $\mathbb C$, or some other topological field, you might want to work with continuous linear maps between topological vector spaces.
Given a linear operator $A: V\to W$, we can define a dual map $A^*: W^* \to V^*$ by $(A^*(\phi))(v)=\phi(A(v))$. It is straight forward to verify that this gives a well defined linear map between the vector spaces. This dual map is the adjoint of $A$. For most sensible choices of dual topologies, this map should also be continuous.
The question is, how does this relate to what you are doing with inner products? Giving an inner product on $V$ is the same as giving an isomorphism between $V$ and $V^*$ as follows:
Given an inner product, $\langle x, y \rangle$, we can define an isomorphism $V\to V^*$ via $x\mapsto \langle x, - \rangle$. This will be an isomorphism by nondegeneracy. Similarly, given an isomorphism $\phi:V\to V^*$, we can define an inner product by $\langle x,y\rangle =\phi(x)(y)$. The "inner products" coming from isomorphisms will not in general be symmetric, and so they are better called bilinear forms, but we don't need to concern ourselves with this difference.
So let $\langle x,y \rangle$ be an inner product on $V$, and let $\varphi$ be the corresponding isomorphism $\varphi:V\to V^*$ defined above. Then given $A:V\to V$, we have a dual map $A^*:V^* \to V^*$. However, we can use our isomorphism to define a different dual map (also denoted $A^*$, but which we will denote by $A^{\dagger}$ to prevent confusion) by $A^{\dagger}(v)=\varphi^{-1}(A^*\phi(v))$. This is the adjoint that you are using.
Let us see why. In what follows, $x\in V, f\in V^*$. Note that $\langle x, \varphi^{-1} f \rangle = f(x)$ and so we have
$$ \langle Ax, \varphi^{-1}f \rangle = f(Ax)=(A^*f)(x)=\langle x, \varphi^{-1}(A^* f) \rangle $$
Now, let $y=\varphi^{-1}f$ so that $\varphi(y)=f$ Then we can rewrite the first and last terms of the above equality as
$$\langle Ax, y \rangle = \langle x, \varphi^{-1}(A^* \phi(y)) \rangle = \langle x, A^{\dagger}y \rangle $$
Best Answer
Let's assume by contradiction that $D$ has an adjoint. In particular, if we set $u = D^{*}(1)$ then we must have the identity $$ f(1) - f(0) = \int_0^1 f'(x) \, dx = \left< Df, 1 \right> = \left< f, D^{*}(1) \right> = \left< f, u \right> $$ for all $f \in \mathbb{R}[x]$. Taking absolute value and using the Cauchy-Schwarz inequality, we get that $$ |f(1) - f(0)| \leq \| f \| \cdot \| u \| = \left( \int_0^1 f^2(x) \, dx \right)^{\frac{1}{2}} \|u\|. $$ Now by plugging $f(x) = x^n$ for $n > 0$, we get that $$ 1 \leq \frac{\|u\|}{\sqrt{2n+1}} $$ for all $n > 0$. This implies that $\sqrt{2n+1} \leq \| u \|$ for all $n > 0$ which is absurd.