For $U \subset V$, we have a natural (continuous) injection
$$\iota^U_V \colon \mathscr{D}(U) \hookrightarrow \mathscr{D}(V).$$
Its transpose,
$$\rho^V_U \colon \mathscr{D}'(V) \to \mathscr{D}'(U)$$
is called the restriction of the distributions on $V$ to distributions on $U$. For regular distributions, that corresponds to the restriction of the locally integrable function representing the distribution to $U$.
The fundamental lemma, that for any $f\in L^1_\text{loc} \Omega$ and $W\subset \Omega$ we have
$$\bigl(\forall \varphi \in \mathscr{D}(W)\bigr)\left(\int_W f(x)\varphi(x)\,dx = 0\right) \implies f\lvert_W = 0 \text{ a.e.},$$
makes the two restrictions compatible.
So here, looking at only test functions with support in $\Omega\setminus\{0\}$ yields the result that $f$ would be zero almost everywhere in $\Omega\setminus\{0\}$, hence almost everywhere in $\Omega$.
For an alternative proof that the Dirac distribution is not regular, consider a test function $\varphi\in\mathscr{D}(\Omega)$ with $\varphi(0) = 1$, and note that then
$$1 = \int_\Omega f(x)\varphi(nx)\,dx$$
for all $n\in\mathbb{N}\setminus\{0\}$ violates the dominated convergence theorem.
Going to take a stab at it, based on http://en.wikipedia.org/wiki/Distribution_%28mathematics%29#Functions_and_measures_as_distributions
By the definition, $$\langle \delta\ast f,\varphi\rangle = \delta(\tilde{f}\ast\varphi) = \delta\left(t\mapsto\int_{\mathbb{R}}\tilde{f}(\tau)\varphi(t-\tau)\,d\tau\right) = \int_{\mathbb{R}}f(-\tau)\varphi(-\tau)\,d\tau = \int_{\mathbb{R}}f(\tau)\varphi(\tau)\,d\tau = \langle f, \varphi\rangle,$$
since integrating over the whole real line backwards is the same as going forwards. Now, isn't there some kind of identification of functions $f$ with distributions $T_f$? Since $\langle \delta\ast f,\varphi\rangle = \langle f,\varphi\rangle$ for all $\varphi$, we see that the distributions $\delta\ast f$ and $f$ are equal.
Best Answer
Suppose there exists $f \in L^1_{\text{loc}}$ such that $T_f = \delta$. Define
$$\varphi_k(x) := \begin{cases} \exp \left(- \frac{1}{1-|kx|^2} \right) & |x| < \frac{1}{k} \\0 & \text{otherwise} \end{cases}.$$
for $k \in \mathbb{N}$. As $\varphi_k \in D$, we get
$$\int \varphi_k(x) f(x) \, dx = T_f(\varphi_k) = \delta(\varphi_k) = \varphi_k(0)= e^{-1}$$
for all $k \in \mathbb{N}$. Hence,
$$e^{-1} = \left| \int \varphi_k(x) f(x) \, dx \right| \leq \underbrace{\|\varphi_k\|_{\infty}}_{\leq e^{-1}} \int_{B(0,1/k)} |f(x)| \,dx.$$
Since $f$ is locally integrable, the dominated convergence theorem yields
$$e^{-1} \leq e^{-1} \inf_{k \in \mathbb{N}} \int_{B(0,1/k)} |f(x)| \, dx =0.$$