For a distribution $g$ in general, the expression $\int_a^b g(x)\,\varphi(x)\,\mathrm d x$ has no meaning in general. It could be interpreted as $\langle g, \varphi\,\mathbf 1_{[a,b]}\rangle$ but then the test functions $\varphi\,\mathbf 1_{[a,b]}$ are not continuous in general.
In the particular case of the Dirac delta however, it can be defined in several ways, i.e. as a measure (see e.g. here). Both coincide when the test functions are smooth with compact support. Any distribution $\mu$ of order $0$ is a measure, and for such a measure one can define the integral with respect to it which verifies
$$
\langle\mu,\varphi\rangle = \int \varphi(x)\,\mu(\mathrm d x)
$$
for any $\varphi$ that is measurable with respect to $\mu$. In the particular case of the Dirac delta, it correspond to ask that $\varphi$ has a well defined value at $0$. This is in contrast with the case of the Lebesgue measure where integrable functions are only defined almost everywhere. Hence, for the Dirac delta, $\mathbf 1_{[0,b]}$ is different from $\mathbf 1_{(0,b]}$, since one has value $1$ at $0$ and the other has value $0$. Hence
$$
\int_{(0,b]} \varphi(x)\,\delta(\mathrm d x) = \langle \delta,\varphi\,\mathbf 1_{(0,b]}\rangle = \varphi(0)\,\mathbf 1_{(0,b]}(0) = 0
$$
while
$$
\int_{[0,b]} \varphi(x)\,\delta(\mathrm d x) = \langle \delta,\varphi\,\mathbf 1_{[0,b]}\rangle = \varphi(0)\,\mathbf 1_{[0,b]}(0) = \varphi(0).
$$
More generally, for distributions of order $1$ (that is measures) one can split integrals in the following ways
$$
\begin{align*}
\int_{[a,b]} \mu(\mathrm d x) &= \int_{[a,c)} \mu(\mathrm d x) + \int_{[c,b]} \mu(\mathrm d x)
\\
&= \int_{[a,c]} \mu(\mathrm d x) + \int_{(c,b]} \mu(\mathrm d x)
\\
&= \int_{[a,c)} \mu(\mathrm d x) + \int_{(c,b]} \mu(\mathrm d x) + \mu(\{c\})
\end{align*}
$$
where $\mu(\{c\})$ is the measure of the set $\{c\}$ by $\mu$. In the case of the Lebesgue measure, a singleton is always of measure $0$ so $\int_{[a,b]}$ and $\int_{(a,b)}$ give the same result, denoted as $\int_a^b$. In the case of the Dirac delta, $\delta(\{0\}) = 1$.