Laurent series and Laurent polynomials
Actually, by using finite $X$ and $Y$, what you've written is only a definition for Laurent polynomial arithmetic, and not Laurent series arithmetic.
The expression you gave carry over directly, but the condition on $X$ and $Y$ is that they have a least element. Notice that if both have a least element, then so do $X\cup Y$ and $X+Y$.
Now for your question
The expression $p = \sum_{i \in X} a_i t^i$ is one way to express $p$, but it's not the only way to write $p$.
If $0\notin X$, then notice that it can be added in for free:
$$p = \sum_{i \in X} a_i t^i=0t^0+\sum_{i \in X} a_i t^i=\sum_{i \in X'} a_i t^i$$ where $X'=X\cup\{0\}$. You can see that $X'$ still has a least element, which is either the least element of $X$ or $0$ itself. Naturally $a_0=0$ in this case. In other cases, you already have $a_0\neq 0$ given to you.
So actually you can see that every coefficient in a Laurent series is defined, it's just that there are limitations that make many of the coefficients $0$, and hence omissible. Such conditions are necessary because they ensure that each coefficient of the sum and product can be obtained from computations with finitely many nonzero numbers.
Identity
I'd like to adapt the notation above so it's a little easier to write things. When I wrote $p_0$, I mean the coefficient of $0$ in $p$. Similarly $q_5$ is the $5$ coefficient of $q$. And $(pq)_{-1}$ is the $-1$ coefficient of the product $pq$.
Using this notation, your arithmetic formulas change into:
$$(p+q)_k = p_k + q_k\\
(p\cdot q)_k = \sum_{i+j=k} p_i q_j$$
The identity is the obvious candidate: the coefficent of $t^0$ should be $1$, and all other coefficients should be $0$. I'm going to call that particular series "$1$".
Checking the multiplication:
$$(p\cdot 1)_k = \sum_{i+j=k} p_i 1_j=p_k$$ (Can you see how that works?) This means that $p\cdot 1=p$, since all of their coefficients match. Thus $1$ is the identity.
Units of $F[[t]]$
The units are not defined as you described. What you wrote was more like a description of the units of $F[[t]]$ (but not a definition.) Let's make that clear here.
To find all the units for the ring of power series over a ring $F$, I'm going to first have you look at units in the power series ring $F[[t]]$, which as you know looks just like $F((t))$ except coefficients below $0$ are always $0$.
First, let's look at units in $F[[x]]$. By definition, $p$ is a unit if there is a $q$ such that $p\cdot q=1$. Suppose $pq=1$. Then the formula for multiplication says that $p_0q_0=1\in F$, so we know for sure that $p_0$ has to be a unit.
Convesely, consider what $q$ must look like to invert $p$ if $p_0$ is a unit in $F$.
We have: $(p\cdot q)_0=p_0q_0=1$ and solving for $q_0$ we get $q_0=p_0^{-1}$. This is possible to write since $p_0$ is a unit of $F$.
Next $(p\cdot q)_1=p_1q_0+p_0q_1=0$, and solving for $q_1$ we get $q_1=-p_1q_0p_0^{-1}$
Next $(p\cdot q)_2=p_2q_0+p_1q_1+p_0q_2=0$, and solving for $q_2$ we get $q_2=-p_0^{-1}(p_2q_0+p_1q_1)$.
Continuing inductively, we produce an infinite series of $q_i$ with the property that their Laurent series $q$ satisfies $p\cdot q=1$. We have now established that units of $F[[t]]$ are exactly the series with a unit of $F$ in first coefficient.
Units of $F((t))$
Finally, we tackle your unit question. First, notice that $t$ is a unit in this ring since it contains the series $t^{-1}$ which is $1$ in the $-1$ spot and $0$ elsewhere. Multiplying you'll find that $t\cdot t^{-1}=1$.
Now let $p$ be any nonzero Laurent series. Obviously $p=t^{z}p'$ where $p'$ is a power series with a nonzero first coefficent. Since $t^z$ is obviously a unit, $p$ is a unit of $F((t))$ iff $p'$ is a unit of $F[[t]]$, but we already know that means that the lowest nonzero coefficient of $p'$ is a unit of $F$.
So there you have it: the units of $F((t))$ are the ones with lowest nonzero coefficient a unit of $F$. If $F$ is a field, then actually $F((t))$ is a field too. It turns out to be the ring of fractions of $F[[t]]$, in that case.
Best Answer
We’re supposing that $I$ is an ideal in $F[[t]]$. For each $p(t)=\sum_{k\ge 0}a_kt^k\in I$, let $$a_p=\min\{k:a_k\ne 0\}\;,$$ so that
$$p(t)=a_{a_p}t^{a_p}+a_{a_p+1}t^{a_p+1}+\ldots=t^{a_p}\left(\underbrace{a_{a_p}+a_{a_p+1}t+a_{a_p+2}t^2+\ldots}_{q(t)\in F[[t]]}\right)\;.\tag{1}$$
Among all elements of $I$, choose $p\in I$ so that $a_p$ is as small as possible, and let $a=a_p$. $(1)$ shows that there is a $q(t)\in F[[t]]$ with a non-zero constant term such that $p(t)=t^aq(t)$, and the claim is that $I=(t^a)$.
Since $q(t)$ has a non-zero constant term, $q(t)$ is a unit in $F[[t]]$, and therefore $t^a\in I$; clearly this implies that $(t^a)\subseteq I$.
Now suppose that $r(t)\in I$, say $r(t)=\sum_{k\ge 0}b_kt^k$. Recall that $p$ was chosen so that $a_p$, the exponent on the first non-zero term of $p(t)$ was as small as possible for any member of $I$. That means that $a_r\ge a_p=a$. But $a_r$ is the exponent on the first non-zero term of $r(t)$, so $b_k=0$ for all $k<a_r$. And since $a\le a_r$, clearly $b_k=0$ for all $k<a$. But that means that every non-zero term of $r(t)$ has an exponent of $a$ or more, which means that we can factor out $t^a$: there is some $s(t)\in F[[t]]$ such that $r(t)=t^as(t)$. Of course this means that $r(t)\in(t^a)$, so we’ve now shown that $I\subseteq(t^a)$.
Putting the two pieces together, we get $I=(t^a)$. Thus, every ideal of $F[[t]]$ is of this form for some $a$.