for the Laplace equation in half-strip where it says "with g(x)=0 we could reduce it by the method of continuation to the problem in the whole strip and solve it by Fourier transform" would that be using a Fourier transform with respect to y this time . . . if yes then would the fact that we would be using an odd continuation influence the answer compared to if we had $u_y|_{y=0}=0$ I don't see how this would influence the solution . . .

Yes, for $u|_{y=0}=0$ we use an odd continuation and for $u_y|_{y=0}=0$ an even continuation. F.e. consider

\begin{align*}

& u_{xx}+u_{yy}=0,\qquad 0<x<1, y>0\\

& u|_{x=0}=0,\\

& u|_{x=1}=e^{-y},\\

&u|_{y=0}=0.

\end{align*}

Then taking an odd continuation $\Phi(y)$ of $\phi(y)=e^{-y}$ which will be $\pm e^{-|y|}$ as $\pm y>0$, we find its FT $ \hat{\Phi}(\eta)=\frac{i\eta}{\pi(1+\eta^2)}$. If instead we had $u_y|_{y=0}=0$ we would take an even continuation and FT of $\Phi$ would be $ \hat{\Phi}(\eta)=\frac{1}{\pi(1+\eta^2)} $.

The rest of solution would be the same (but with different $\hat{\Phi}(\eta)$ !):

\begin{gather*}

\hat{u}_{xx}-\eta^2 \hat{u}=0 \implies \hat{u}= A(\eta)\cosh (x\eta) + B(\eta)\sinh(x\eta),\\

\hat{u}|_{x=0}=0\implies A(\eta)=0 ,\qquad \hat{u}|_{x=1}=\hat{\Phi}(\eta)\implies B(\eta)= \hat{\Phi}(\eta)/\sinh (\eta),\\

\hat{u}= \hat{\Phi}(\eta)\sin(x\eta)/\sinh (\eta)\implies \\

u(x,y)= \int_{-\infty}^\infty \frac{\hat{\Phi}(\eta)\sin(x\eta)}{\sinh (\eta)} e^{iy\eta}\,d\eta.

\end{gather*}