How to show that the integral is finite for $(x,t)$ such that $x>0$ and $t>0$?

Hint. Since $g:[0,+\infty) \rightarrow R$ is continuous and bounded, the potential problems are near $s=0^+$ and near $s=t^-$.

1. For $s$ near $0^+$, we have $$ \frac{1}{(t-s)^{3/2}}e^{\frac{-x^2}{4(t-s)}}g(s) \sim \frac{1}{t^{3/2}}e^{\frac{-x^2}{4t}}g(0) $$ and the integral '$ \displaystyle \int_{0}\frac{1}{t^{3/2}}e^{\frac{-x^2}{4t}}g(0)\:ds $' is finite.
2. For $s$ near $t^-$, we have $$ \frac{1}{(t-s)^{3/2}}e^{\frac{-x^2}{4(t-s)}}g(s) \sim \frac{1}{(t-s)^{3/2}}e^{\frac{-x^2}{4(t-s)}}g(t) $$ and the integral '$ \displaystyle \int^{t}\frac{1}{(t-s)^{3/2}}e^{\frac{-x^2}{4(t-s)}}\:ds $' is finite, using for example the fact that $$ 0<\frac{1}{(t-s)^{3/2}}e^{\frac{-x^2}{4(t-s)}}< \frac{1}{(t-s)^{1/2}} $$ for $s$ sufficiently near $t^-$.

An elegant way to show that the integral is finite, is to change your integration variable from $s$ to $q = \frac 1{t-s}$. Note that $ds = dq (t-s)^2$, so you get rid of the only term that might cause divergence: $(t-s)^{-3/2}$. The resulting integral looks neat.