First note that
$$ \partial_x k(s, x) = -\frac{x}{s}k(s,x),
\quad s > 0, x \in {\mathbb R},$$
so your integral becomes
$$
q_a(t,x) := \int^t_0 \partial_x k(s,x-a)k(t-s,a)ds.
$$
Now suddenly your integral $q_a(t,x)$ becomes a representation of the unique classical solution to the boundary value problem for the heat equation (see e.g Cannon's book One-Dimensional Heat Equation, Ch 4):
\begin{align}
\partial_x q_a(t,x) &= \frac{1}{2} q_a(t,x), \\
q_a(0,x) &= 0, \quad x < a, \\
q_a(t,a) &= k(t,a) \quad t > 0.
\end{align}
The constant boundary $x = a$ is regular for this problem, and the solution achieves the boundary values continuously, i.e.,
$$
\lim_{x \nearrow a} q_a(t,x) = k(t,a).
$$