The formula will work just fine (with no pseudo required), provided you interpret $\widehat{f}$ suitably. Since $f$ is only a function on a half space, we have to make an agreement on what exactly we mean by $\widehat{f}$.

To obtain the Dirichlet Laplacian, define $\widehat{f}$ as the Fourier transform of the odd extension $f(x,-y)=-f(x,y)$ of $f$ (I use the notation $x=(x_1,\ldots ,x_{n-1})$, $y=x_n$). Then $$ \widehat{-\Delta f} = |\xi|^2 \widehat{f} , $$ as expected. Moreover, the domains also get handled correctly by this formula: $f\in D(-\Delta)=W^2_0$ precisely if the odd extension of $f$ satisfies $|\xi|^2\widehat{f}\in L^2(\mathbb R^n)$, or, equivalently, if (in $L^2$ sense) $$ f(x,y) = \int_{\mathbb R^{n-1}} dx\int_0^{\infty} dy\, g(\xi',k) e^{i\xi' x}\sin ky $$ for some $g\in L^2$ with $(|\xi'|^2+y^2)g\in L^2$.

For the Neumann Laplacian, work with even extensions instead.

The formula will work just fine (with no pseudo required), provided you interpret $\widehat{f}$ suitably. Since $f$ is only a function on a half space, we have to make an agreement on what exactly we mean by $\widehat{f}$.

To obtain the Dirichlet Laplacian, define $\widehat{f}$ as the Fourier transform of the odd extension $f(x,-y)=-f(x,y)$ of $f$ (I use the notation $x=(x_1,\ldots ,x_{n-1})$, $y=x_n$). Then $$ \widehat{-\Delta f} = |\xi|^2 \widehat{f} , $$ as expected. Moreover, the domains also get handled correctly by this formula: $f\in D(-\Delta)=W^2_0$ precisely if the odd extension of $f$ satisfies $|\xi|^2\widehat{f}\in L^2(\mathbb R^n)$, or, equivalently, if (in $L^2$ sense) $$ f(x,y) = \int_{\mathbb R^{n-1}} d\xi' \int_0^{\infty} dk\, g(\xi',k) e^{i\xi' x}\sin ky $$ for some $g\in L^2$ with $(|\xi'|^2+y^2)g\in L^2$.

For the Neumann Laplacian, work with even extensions instead.