Let $A, B$ be unbounded self-adjoint operators on Hilbert spaces $\mathcal{H_1}, \mathcal{H_2}$ with spectra $\sigma(A), \sigma(B)$. We know that : $\sigma(A \otimes 1 + 1 \otimes B) = \overline{\sigma(A) +\sigma(B)}$ and $A \otimes 1 + 1 \otimes B$ defines an unbounded self-adjoint operator on $\mathcal{H_1} \otimes \mathcal{H_2}$ (after taking its closure). By the spectral theorem : $A = \int_{\sigma(A)} \lambda \,dE_A(\lambda)$; $B = \int_{\sigma(B)} \lambda\, dE_B(\lambda)$; $A \otimes 1 + 1 \otimes B = \int_{\overline{\sigma(A) +\sigma(B)}} \lambda\, dE_{A \otimes 1 + 1 \otimes B}(\lambda)$.
My question is : do we have $E_{A \otimes 1 + 1 \otimes B} = E_A \otimes E_B$ ? If so, how does one show this, do you have a reference that I could cite ? I guess one would have to state that there exists a common spectral family, then taking the closure of $A \otimes 1 + 1 \otimes B$ would yield integrating over the closure of the spectrum, but it seems touchy so I'm looking for a reference.
While the approach from the comments using multiplication operators is probably the standard way to go and also easier, I like this approach using the Fourier inversion formula and unitary groups.
Since $(e^{itA}\otimes e^{itB})$ is a strongly continuous group of unitaries, there exists a self-adjoint operator $H$ such that $e^{itA}\otimes e^{itB}=e^{itH}$ by Stone's theorem. If $\xi\in D(A)$ and $\eta\in D(B)$, then \begin{align*} \frac 1 t(e^{itA}\otimes e^{itB}(\xi\otimes \eta)-\xi\otimes\eta)&=\frac 1 t(e^{itA}\xi-\xi)\otimes e^{itB}\eta+\frac 1 t\xi\otimes (e^{itB}\eta-\eta)\\ &\to A\xi\otimes \eta+\xi\otimes B\eta. \end{align*} Thus $A\odot 1+1\odot B\subset H$, which implies $A\otimes 1+1\otimes B=H$ (you cannot have proper inclusions of self-adjoint operators).
Let $E$ denote the spectral measure of $H$. If $\phi\in\mathcal S(\mathbb R)$ and $\xi\in \mathcal H_1\otimes\mathcal H_2$, then \begin{align*} \int \phi(t)\,d\mu_\xi(t)&=\int\hat\phi(t)\,d\check\mu_\xi(t)\\ &=\int\int\hat\phi(t)e^{2\pi it\lambda}\,d\mu_\xi(\lambda)\,dt\\ &=\int\hat\phi(t)\langle\xi,e^{2\pi itH}\xi\rangle\,dt\\ &=\int\hat\phi(t)\langle \xi,e^{2\pi it A}\otimes e^{2\pi it B}\xi\rangle\,dt\\ &=\int\langle\xi,\int\int\hat\phi(t)e^{2\pi i(\lambda+\mu)t}\,d(E_A\otimes E_B)(\lambda,\mu)\xi\rangle\,dt\\ &=\int\int \hat\phi(t)e^{2\pi i(\lambda+\mu)t}\,dt\,d\langle\xi,(E_A\otimes E_B)\xi\rangle(\lambda,\mu)\\ &=\int \phi(\lambda+\mu)\,d\langle\xi,(E_A\otimes E_B)\xi\rangle(\lambda,\mu). \end{align*} It follows by approximation that $$ \mu_\xi(S)=\int 1_S(\lambda+\mu)\,d\langle\xi,(E_A\otimes E_B)\xi\rangle(\lambda,\mu) $$ for every Borel set $S\subset \mathbb R$.