Already in an arbitrary Hilbert space, the product of two unbounded operators need not be defined. The reason is that unbounded operators usually are not defined on the whole Hilbert space but only on a domain dense in the Hilbert space. The product $AB$ is another operator only if there is a dense subspace of the domain of $B$ that is mapped by $B$ into the domain of $A$. This is a nontrivial condition that must be checked in each case. In particular, this holds for Fock space, the arena of free quantum fields.

Now quantum fields are not even operators in Fock space, only operator-valued distributions. In particular, the number operators $N(x):=a^*(x)a(x)$ are well-defined only in a distributional sense. Meaningful as densely defined operators (with a common domain) are only the integrals $N(f):=\int dx f(x) N(x)$ with a Schwarz test function $f$. Thus products of the form $N(f)N(g)$ can be proved to be well-defined on a suitable domain, while the product $N(x)N(y)$ is questionable.

In nonrelativistic QFT, this proves to be harmless as products of number operators appear only in integrated form, and the integral is nice enough to make operators such as the potential energy well-defined. However, in relativistic QFT, the integral is a local integral, which is not well-defined without renormalization.