Generally, given $T\colon \mathcal{D}(T) \to H_2$, where $H_1, H_2$ are Hilbert spaces, $\mathcal{D}(T)$ is a dense subspace of $H_1$, and $T$ is a linear operator, the adjoint of $T$ is defined on the subspace $\mathcal{D}(T^\ast) \subset H_2$ of elements $y$ such that there exists a $z\in H_1$ with $\langle Tx,y\rangle_2 = \langle x, z\rangle_1$ for all $x\in \mathcal{D}(T)$, then $T^\ast(y) = z$. In short,
$$\langle Tx,y\rangle_2 = \langle x, T^\ast y\rangle_1$$
for all $x\in \mathcal{D}(T),\, y \in \mathcal{D}(T^\ast)$. The denseness of $\mathcal{D}(T)$ ensures the well-definedness of $T^\ast$.
In the situation at hand, the codomain of $T$ is $\mathbb{K}$ (whether that's $\mathbb{R}$ or $\mathbb{C}$ doesn't matter), so there are two possibilities for $\mathcal{D}(T^\ast)$; it can be either $\{0\}$ or $\mathbb{K}$. If $\mathcal{D}(T^\ast) = \mathbb{K}$, then $T^\ast\colon \mathbb{K}\to L^2$ is continuous, and hence it has a continuous adjoint $T^{\ast\ast}\colon L^2 \to \mathbb{K}$. But we then have $T \subset T^{\ast\ast}$, so $T$ itself would be continuous. The given $T\colon f \mapsto f(0)$ is not continuous, hence $\mathcal{D}(T^\ast) = \{0\}$, i.e. $T^\ast$ is not densely defined.
The argument shows, with minor modifications, that a densely defined operator with finite-dimensional codomain has a densely defined adjoint if and only if it is continuous, since the only dense subspace of a finite-dimensional Hausdorff topological vector space is the entire space, and every linear operator $\mathbb{K}^n \to V$, where $V$ is a topological vector space, is continuous.
First of all, the condition you state above does NOT imply that $T$ is bounded. Moreover, Heilinger Toeplitz can only be applied if $T$ is defined on all of $H$, i.e. $\mathcal{D}(T)=H$, which is not one of your assumptions.
However, your assumption does indeed imply that $T$ is a (possibly unbounded) densely defined closable operator. To see this, your assumption implies that the numerical range $W(T)$ of $T$, defined as
$$
W(T):=\{(Tx, x) \in \mathbb{C} \mid x \in \mathcal{D}(T),\,\|x\|=1\},
$$
(I use the convention that the inner product is linear in the first argument) is contained in the closed right half-plane. That means, in particular, $W(T)\neq \mathbb{C}$. All densely defined operators satisfying this must be closable. This is a classical result which can be seen for instance in Chapter V Thm. 3.4. of Kato, T. Perturbation theory for linear operators.
By the way, the condition you state above is called accretive.
Best Answer
Condition $(B)$ is equivalent to condition $(A)$. In this answer I show that $(B) \implies (A)$ since you know how to do the other direction.
The key point is that a closed subspace $\mathcal{G} \subset H \oplus H$ is the graph of a closed extension of $A$ if and only if $\mathcal{G}(A) \subset \mathcal{G}$ and $\mathcal{G}$ has the single valued property $$(x,y_1),(x,y_2) \in \mathcal{G} \implies y_1 = y_2$$ which is equivalent to $$(0,y) \in \mathcal{G} \implies y = 0$$ since $\mathcal{G}$ is a linear space. In general, $K \subset H \oplus H$ is the graph of an operator if and only if it is a linear subspace with the single valued property.
So we want to check that if $(B)$ holds then $\overline{\mathcal{G}(A)}$ has the single valued property. But this is easy since if $(0,y) \in \overline{\mathcal{G}(A)}$ then $(0,y) \in \mathcal{G}(\tilde{D})$ since $\mathcal{G}(\tilde{D})$ is a closed subset containing $\mathcal{G}(A)$. Since $\mathcal{G}(\tilde{D})$ is the graph of an operator, it has the single valued property and so $y = 0$. Hence $\overline{G(A)}$ has the single valued property and so is the graph of an operator.