I read that $\mathbb{Z}$ is not a complete lattice because it has no greatest lower bound (and lub).
If we define $-\infty$, $\infty$ to be integers, i.e., include them in $\mathbb{Z}$, does $\mathbb{Z}$ become a complete lattice?
If yes, then does it seem absurd to not call $\mathbb{Z}$ a complete lattice just because of something as arbitrary as inclusion of infinities in the set of integers, which is a matter of opinion?
Yes, $(\mathbb{Z}\cup\{-\infty,\infty\},\leq)$ is a complete lattice, as given $S\subset \mathbb{Z}\cup\{-\infty,\infty\}$, either $S$ is bounded above, in which case the greatest integer in $S$ is the supremum, or its supremum is $\infty$, and likewise either $S$ is bounded below, in which case the least integer in $S$ is the infimum, or its infimum is $-\infty$.
What complete means is not really a matter of opinion. A lattice $(L,\leq)$ is complete exactly when every subset of $L$ has a supremum and infimum, and since $L$ and $\emptyset$ are perfectly good subsets of $L$, we know that their supremum and infimum must exist, and this provides the minimum and maximum of $L$.
To perhaps capture what you want, however, is the idea of Dedekind completeness. I.e. a lattice $(L,\leq)$ is said to be Dedekind complete exactly when every non-empty subset $S$ of $L$ which has an upper bound (i.e. bounded above) has a least upper bound. By adding a maximum and minimum to a Dedekind complete lattice, you can make a complete lattice (though the reverse does not hold; you can't in general take a complete lattice and remove its minimum and maximum and have a Dedekind complete lattice).
This is in contrast to a bounded-complete lattice, in which the above non-emptiness condition is not included (and so $\emptyset$ has a supremum, which is necessarily a minimum).