The Fourier transform is a unitary operator from $L^2$ to $L^2$. But all infinite-dimensional Hilbert spaces are isometrically isomorphic to $L^2$. So that means we can define the Fourier transform on arbitrary infinite-dimensional Hilbert spaces.
So my question is, what does the Fourier transform look like on an arbitrary infinite-dimensional Hilbert space? Can it still be defined in terms of an integral?
There is nothing as the Fourier transform in arbitrary Hilbert space. The Fourier transform is the unitary operator diagonalizing the shift operator $S_a f(x) = f(x+a)$ and every convolution operator.
That is it applies to $L^2(G)$ for $G$ an abelian group, and it is unitary $L^2(G) \to L^2(\widehat{G})$ the Pontryagin dual.
With $G = \Bbb{R}$ then $\widehat{G} = \Bbb{R}$.
With $G = \Bbb{R/Z}$ then $\widehat{G} = \Bbb{Z}$, the corresponding Fourier transform is called Fourier series.