You can use the following (weak) argumentation to deduce that the result is interpreted as the (not very meaningful statement of)
$$ \delta(t) \cdot \delta(t) = \delta(0) \cdot \delta(t) $$
Define the impulse as a limit of the following pulse function
$$ \delta(t) = \lim_{\Delta \to 0} \delta_{\Delta}(t) $$ where $\delta_{\Delta}(t)$ is a rectangular function with the definition that
$$ \delta_{\Delta}(t) = \begin{cases}
0 &,\ t < 0 \\
\frac{1}{\Delta} &,\ 0 \le t < \Delta \\
0 &,\ \Delta \le t
\end{cases} $$
Then you can define the multiplication as:
$$\begin{align}
\delta(t)\delta(t) &= \left( \lim_{\Delta \to 0}\delta_{\Delta}(t) \right) \left( \lim_{\sigma \to 0}\delta_{\sigma}(t) \right)\\
&= \lim_{\Delta \to 0} \lim_{\sigma \to 0} \left[ \delta_{\Delta}(t) \delta_{\sigma}(t) \right]\\
&= \lim_{\Delta \to 0} \left( \delta_{\Delta}(t) \lim_{\sigma \to 0} \delta_{\sigma}(t) \right)\\
\text{now observe that}\\
\delta_{\Delta}(t) \lim_{\sigma \to 0} \delta_{\sigma}(t) &\approx \delta_{\Delta}(0) \lim_{\sigma \to 0} \delta_{\sigma}(t) = \delta_{\Delta}(0) \delta(t)
\end{align} $$ where the equality holds in the limit. Then proceed with
$$
\begin{align}
\delta(t)\delta(t) &= \lim_{\Delta \to 0} \left( \delta_{\Delta}(0) \delta(t) \right) \\
\delta(t)\delta(t) &= \left( \lim_{\Delta \to 0} \delta_{\Delta}(0) \right) \delta(t) \\
\delta(t)\delta(t) &= \left( \delta(t)|_{t=0} \right) \delta(t) \\
\delta(t)\delta(t) &= \delta(0) \delta(t) \\
\end{align}
$$