As the title says, I'm trying to solve the wave equation with the Laplace transform (withe respect to t) so
$$u_{tt}-c^{2}u_{xx}=0$$
with the conditions $u(x,0)=u_{t}(x,0)=0$ and $u_{t}(0,t)=g(t)$ where $G(p)=L[g]$.
I can find the transforms of the differentials and substitute in the first two conditions to reach
$$p^{2}U(x,p)-c^{2}\frac{\partial^{2}U}{\partial x^{2}}=0$$
which I can then solve. I then remove the $A(p)e^{\frac{px}{c}}$ term from consideration because we do not wish for our solution to diverge as $x \to \infty$ which leaves:
$$U(x,p)=B(p)e^{\frac{-px}{c}}$$
I need to apply the last condition to this however I don't know how to do so?
You have $u_t(0,t)=g(t)$, so this means $$L[u_t](0,p)\color{red}=pL[u](0,p)-u(0,0)=L[g](p)\\\implies pU(0,p)-0=G(p)\\\implies U(x,p)=G(p)\frac{e^{-px/c}}p$$ Given a function $g(t)$, you can then invert this to get a solution in terms of $x$ and $t$. This can be done using the convolution theorem as follows.
Note that the Laplace transform of the Heaviside Step function is $$L[\theta(t)]=\frac1p$$This means that the Laplace transform of $\theta[at-1]$ for some positive constant $a$ is $$L[\theta(at-1)]=\frac1p e^{-p/a}$$ You can use this result to compute the inverse Laplace transform of the other term in your expression of $U$. Denote this function $f(t)$. Then convolution theorem tells us $$L[f(t)*g(t)]=L[f(t)]L[g(t)]$$So $$u(x,t)=g(t)*f(t)$$The fact that there is a Heaviside function involved will likely alter the limits to give what you have there.
The $\color{red}{\text{red}}$ step was done using the property that $$L[f'(t)]=pL[f(t)]-f(0)$$