Your claim is true, here goes the proof following your main intuition. In the following WLOG $x=0$. Consider a compactly supported smooth function $\phi: \mathbb{R} \to [0,1]$ such that $\text{supp}(\phi)\subset B(0,2)$ and $\phi =1 $ on $B(0,1)$, such that $\|\phi'\|_{\infty} \leq 2$. We will use as test functions suitable scaled versions of $\phi$, namely we consider for all $k\in \mathbb{N},k\geq 1$ the function $\phi_k$ defined via $$ \phi_k (x) = \phi(kx) $$ $\forall x\in \mathbb{R}$. Clearly $\text{supp}(\phi_k) \subset B(0, \frac{2}{k})$ and $\phi_k = 1$ on $B(0,\frac{1}{k})$. Notice that for the functions $\phi_k$ it holds that $\phi_k'(x) = k\phi'(kx)$ and thus that $\|\phi_k'\|_\infty \leq 2k$. Now, following your idea, we compute: \begin{align*} \int_\mathbb{R} \phi_k \, dDu &= -\int_{\mathbb{R}} u(y)\phi_k'(y)\,dy \\&= -\int_\mathbb{R} (u(y)-u(0))\phi_k'(y)\,dy. \end{align*} where I have used that $\int_\mathbb{R} u(0) \phi_k'(y) \,dy = 0$ since $\phi_k$ has compact support. We now try to estimate the right-hand side of the previous equation. Since $\text{supp}(\phi_k) \subset B(0,\frac{1}{k})$, and recalling the bound $\|\phi'_k\|_{\infty} \leq 2k$ we have: \begin{align*} \left|\int_\mathbb{R} (u(y)-u(0))\phi'_k(y)\,dy\right| &\leq \int_{-\frac{1}{k}}^{\frac{1}{k}} |u(y)-u(0)||\phi'_k(y)| \, dy \\&\leq \int_{-\frac{1}{k}}^{\frac{1}{k}} |u(y)-u(0)|2k \, dy \\&= 4\frac{1}{\frac{2}{k}} \int_{-\frac{1}{k}}^{\frac{1}{k}} |u(y)-u(0)|\,dy \to_{k\to \infty} 0 \end{align*} where the last passage follows from the assumption that 0 is a Lebesgue point. Therefore we have obtained that $\left|\int_\mathbb{R} \phi_k \,dDu\right| \to 0$ At this stage, de dominated convergence theorem implies $$ 0 = \lim_{k\to \infty} \int_{\mathbb{R}} \phi_k \, dDu = \int_{\mathbb{R}}\chi_{\{0\}} \, dDu = Du(\{0\}). $$ You have just to recall the definition of total variation of a measure $\mu$, that is $$ |\mu(E)| = \sup\{\sum_{i=1}^\infty |\mu(E_i)|\mid \{E_i\}_1^\infty \text{ is a partition of } E\}. $$ Indeed, since the only partition of the singleton $\{0\}$ is $\{\{0\}\}$ you obtain $|Du|(\{0\}) = |Du(\{0\})| = 0$.