The wikipedia page for Categorical theory answers your question (emphasis mine):
Saharon Shelah extended Morley's theorem to uncountable languages: if the language has cardinality $\kappa$ and a theory is categorical in some uncountable cardinal greater than or equal to $\kappa$ then it is categorical in all cardinalities greater than $\kappa$.
Now this is actually a bit stronger than you might expect! Morley's theorem says that if a theory in a countable language is categorical in a cardinal greater than $\aleph_0$, then it is categorical in all cardinalities greater than $\aleph_0$. On the other hand, if a theory in a countable language is categorical in a cardinal equal to $\aleph_0$ (i.e. the theory is countably categorical), this does not guarantee categoricity in any other cardinals.
Shelah's possibly surprising result, as stated above, puts together two theorems from Chapter IX of Classification Theory (pages 490 and 491). Theorem 1.16 is the natural generalization of Morley's theorem, and Theorem 1.19 deals separately with the case of a theory $T$ which is $|T|$-categorical, showing that this case trivializes when $|T|$ is uncountable.
THEOREM 1.16: Suppose $T$ is categorical in some $\lambda > |T|$ or every model of $T$ of cardinality $\lambda$ (for some $\lambda>|T|$) is $|T|^+$-universal. Then $T$ is categorical in every $\mu > |T|$, and every model of $T$ of cardinality $>|T|$ is saturated.
THEOREM 1.19: If $T$ is categorical in $|T|>\aleph_0$, then $T$ is a definitional extension of some $T'\subseteq T$, $|T'|<|T|$.
The point is that if $T$ is a definitional expansion of $T'$, then there is a one-to-one cardinality-preserving correspondence between the models of $T$ and the models of $T'$. If $T$ is $|T|$-categorical, then $T'$ is also $|T|$-categorical. Since $|T'|<|T|$, by Theorem 1.16, $T'$ is $\kappa$-categorical for all $\kappa > |T'|$. So also $T$ is $\kappa$-categorical for all $\kappa>|T'|$, and in particular for all $\kappa>|T|$.
What's happened here is that the uncountable $|T|$-categorical theory $T$ is "secretly" just a theory of smaller cardinality. And this is really quite a silly situation to be in! To say that $T$ is a definitional expansion of $T'$ is to say that every symbol in the language of $T$ which is not in the language of $T'$ is defined by a formula in the language of $T'$. But there are only $|T'|$-many formulas in the language of $T'$, so while there are $|T|$-many new symbols, up to equivalence there are only $|T'|$-many!
To give an explicit example, we could take $T'$ to be the theory of algebraically closed fields and take $T$ to be the theory obtained by introducing uncountably many constant symbols $\{c_\alpha\mid \alpha\in \kappa\}$ and setting them all to $0$ by adding axioms $c_\alpha = 0$ for all $\alpha$. Theorem 1.19 says that every example has to be almost as trivial as this one. Upshot: the behavior of countable $\aleph_0$-categorical theories is very special to the countable.
You write:
An isomorphism $\pi\colon A\to B$ can be 'dumbed down' to 3 main
points
- $A$ and $B$ are elementary equivalent (for $\varphi\in L$ then $M\models \varphi \iff N\models \varphi$)
- $A$ and $B$ have the same 'structure' ($n$-ary functions and $n$-ary relations are preserved)
- $A$ and $B$ are bijective
This "dumbed down" interpretation is imprecise and incorrect. I think it's giving you entirely the wrong intuition.
First: I don't know what is meant by "$A$ and $B$ are bijective". Bijectivity is a property of a function, not a pair of sets. If you mean that $A$ and $B$ have the same cardinality, that's true whenever $A\cong B$. But in order for $\pi\colon A\to B$ to be an isomorphism, you need to know that $\pi$ itself is a bijection, not just that there exists some bijection between $A$ and $B$.
Similarly, the fact that $A$ and $B$ are elementarily equivalent is a consequence of $A\cong B$. But it is not part of the definition of isomorphism, and it is certainly not sufficient for $A\cong B$. It is possible to have structures $A$ and $B$ with $A\equiv B$ (elementarily equivalent) and $|A| = |B|$ (same cardinality), but $A\not\cong B$.
The correct definition is that $\pi\colon A\to B$ is an isomorphism if $\pi$ is a bijection $A\to B$ which preserves the function symbols in $L$ and preserves and reflects the relation symbols in $L$. This is what you need to check to show that two structures are isomorphic, not the "check boxes" you listed.
I have no idea why you think the quote from Wikipedia supports your idea. It just says that if $T$ (in a countable language!) has some infinite model, then it has infinite models of all infinite cardinalities. It does not say anything about isomorphisms between these models. (Put another way: It's saying $I(T,\kappa)\geq 1$, not $I(T,\kappa) = 1$.)
Moral: Don't try to "dumb down" mathematical concepts. Without precise definitions, you won't get anywhere.
Best Answer
You have the right idea to prove that the theory is totally transcendental by proving that it is $\omega$-stable. But this is a non-trivial component of Morley's theorem, and one that you're not going to be able to prove just by playing around with Löwenheim-Skolem.
It's also a step that will be included in any proof of Morley's theorem (e.g. it's Corollary 5.2.10 in Marker, Theorem 5.2.4 in Tent and Ziegler, and Theorem 3.8 in Morley's original paper). Maybe "I am studying Morley's theorem" means that you're trying to reprove the whole thing on your own! But if not, you're probably reading from some reference that includes this proof.
In any case, the idea is this: You show that for every every countable theory $T$ with infinite models and every infinite cardinal $\kappa$, there is a model $M\models T$ which only realizes countably many types over any countable set of parameters. How? You build an Ehrenfeucht-Mostowski model generated by a sequence of indiscernibles ordered like the ordinal $\kappa$ (this is the tricky part of the proof, which would take far too long to explain in this answer). But on the other hand, if there is a countable model $N$ of $T$ such that $|S_1(N)|\geq \aleph_1$, then you can realize a set of $\aleph_1$-many types over $N$ in an elementary extension $N\preceq N'$ of size $\kappa$. These two models $M$ and $N'$ aren't isomorphic, so $T$ isn't $\kappa$-categorical.