Let $\Omega$ denote any set of one or more mutually commuting self-adjoint operators, such as the observables corresponding to the $x$-coordinates of two particles. Let $\Omega'$ denote the commutant, which is the set of all oeprators that commute with everything in $\Omega$. Then let $\Omega''$ denote the double commutant, which is the set of all operators that commute with everything that commutes with everything in $\Omega$. (That's not a typo.) We can also describe $\Omega''$ as the (commutative) von Neumann algebra generated by $\Omega$.
(Technically, a von Neumann algebra contains only bounded operators, and operators like $X$ are unbounded, but that technicality doesn't affect the spirit of this answer.)
If the operators in $\Omega$ qualify as observables in the given model, then the algebra $\Omega''$ contains all of the projection operators that we need to characterize the possible outcomes of a simultaneous measurement of the observables in $\Omega$.
Here's the key: Every commutative von Neumann algebra is generated by a single self-adjoint operator . That proves the existence of a single self-adjoint operator representing the simultaneous measurement of all of the observables in $\Omega$. Actually constructing such an operator is a different problem, and it probably wouldn't be useful. (As explained in knzhou's answer, $X\otimes X$ doesn't work.) Using separate operators, one for each coordinate of each particle, is more convenient.
 EP10 on page 23 in Jones (2009), "Von Neumann Algebras," https://math.berkeley.edu/~vfr/VonNeumann2009.pdf
 Lemma 1 in Suzuki and Saitô (1963), "On the operators which generate continuous von Neumann algebras," https://projecteuclid.org/euclid.tmj/1178243811