= Solution
Both kernels are <integrable functions>, so the <convolution theorem> gives
$$
\widehat{q_t*q_s}(\omega)=\widehat q_t(\omega)\widehat q_s(\omega)
=e^{-(t+s)\omega^2}=\widehat q_{t+s}(\omega).
$$
The <uniqueness of the Fourier transform on integrable functions> gives equality almost everywhere. Both sides are continuous, so it holds at every point:
$$
\boxed{q_t*q_s=q_{t+s}.}
$$
For completeness, continuity of the convolution follows by <dominated convergence>: $q_t$ is bounded and continuous, so for $x_j\to x$ the integrands $q_t(x_j-y)q_s(y)$ converge pointwise and are dominated by $\|q_t\|_\infty q_s(y)$.
The identity is the <heat semigroup> property: evolving for time $s$, then for time $t$, is the same as evolving once for time $s+t$.
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