There are a couple of ways to answer your question. First, the short answer:

The integers can be made into a sequence simply by choosing any first element. For example, @BAI's comment puts the integers in their standard order $0, 1, -1, 2, -2, \ldots$.

A slightly longer answer: Any countably infinite set $A$ may be made into a sequence by induction (or by reference to the bijection with $\Bbb{N}$ that witnesses the countability of the set). One simply chooses a first element $a_1$, then a second element $a_2 \neq a_1$, and so on.

Maybe more interestingly, there are so-called "bi-infinite" sequences, as mentioned here. Such things don't have a first element, and the integers in their natural order form a bi-infinite sequence.

Looking at a set of integers, am I right to say that we cannot have a sequence whose terms are all the integers?

No, that is absolutely wrong.

I mean that {...,−3,−2,−1,0,1,2,3,4,...} is not a sequence because every sequence, finite or infinite, must have the first term.

$\{...,−3,−2,−1,0,1,2,3,4,...\}$ is not a sequence but $\{0,1,-1,2,-2,3,-3,..... \}$ is. It very easy to list all the integers in such a way that you have a first element.

Since a sequence can be thought of as a function whose domain is a set of positive integers? Is that so?

I think this might be the key to your misunderstanding. Yes, the domain is a set of positive integers arranged from smallest to larger one after another. And the range is a set ordered by position one after another. But they needn't be, and usually are not, ordered by size. They only need to be ordered by position.

The first can be anyone you want, the second any other, and you can certainly go through them all even though there is never a smallest one to start with. You don't need to start with the smallest term in the range. You need to start with the first term in the range which could be any size.

It's very possible to have a $f:\mathbb N \to \mathbb Z$. It's even possible for the function to be onto.

It is true $a < b$ will not mean that $f(a) < f(b)$ but there utterly no reason to think it should. And yes it is true that if $f:\mathbb N \to \mathbb Z$ is onto it will be impossible for $a<b$ to mean $f(a) < f(b)$. But that is by no means a requirement.

You could set up e.g. the sequence $\langle 0, 1, -1, 2, -2, 3, -3, \dotsc \rangle$. There are many others (like 0, then five positive numbers, then ten negative, then next five positive, ...).

More deeply: there are sets that can be placed in one-to-one correspondence with the natural numbers $\mathbb{N} = \{1, 2, \dotsc\}$, they are called countable. As above, the integers $\mathbb{Z}$ are a countable set. More interestingly, the set of pairs of integers is also countable (think of the pairs $(i, j)$ laid out in an infinite array, and walk along diagonals: $(1, 1)$, then pairs adding up to 3 $(1, 2), (2, 1)$, then pairs adding to 4, $(1, 3), (2, 2), (3, 1)$, and so on). This means that the set of rational numbers $\mathbb{Q}$ is also countable! But it turns out that the set of real numbers, $\mathbb{R}$, is not countable (one would say it is "larger" in some sense, even if all are infinite).

That depends on what you want to call what.

If you define a sequence to be a map defined on a well-ordered subset of the integers (that is, so that a sequence would always have a "first" member), then fine.

But in some other problems, it may be convenient for you to consider a sequence as being a map on any subset of the integers.

The integers is a countable set, so you can enumerate the terms within the integers as a sequence. The sequence is as follows 0,1,-1,2,-2,3,-3...... and so on.