Given two compatible observables $A$ and $B$ with a common eigenbasis, the completeness relation is: $\newcommand{\ket}[1]{|#1\rangle} \newcommand{\bra}[1]{\langle#1|}$ $$ \sum_{i,j}\ket{a^i,b^j}\bra{a^i,b^j} = 1 $$
Since $\ket{a^i,b^j}$ is not guaranteed to exist for all combinations of $i$ and $j$, does the sum imply we simply ignore the terms which don't exist?
There is no stipulation that the set of $b$’s have same cardinality as the set of $a$’s, and the sum is limited to those sets where so there is no state for which $\vert a^i,b^j\rangle$ exist if $b^j$ is not in the set of $b$ and same for $a^i$. The indices $i$ and $j$ are independent so need not range over the same index set.
Thus, for instance, assuming that $b^i$ takes one of two values in $\{+,-\}$, and $a^i$ are energies of a harmonic oscillator, we would have $$ \sum_{i=0}^\infty\sum_{j=-,+}\vert E_i,j\rangle\langle E_i,j\vert = \sum_{i=0}^\infty\vert E_i,+\rangle\langle E_i,+\vert + \sum_{i=0}^\infty\vert E_i,-\rangle\langle E_i,-\vert=\mathbb{1} $$
