I'm going to start with your example and work towards a more abstract notion of structure throughout this writing. So let's see, the bijection you give is a function $f:A\rightarrow B$. But all we have are the sets $A,B$. No other information is given. So what does the bijection encode? Well, both sets have $3$ elements. Perhaps that is what we should look at. So, let $$M\overset{f}\longrightarrow N$$ be a bijection between sets. If we know $M$ is of finite cardinality, it is not too difficult to deduce from the pigeon hole principle that $N$ is also of finite, equivalent, cardinality. We use this notion for the infinite as well. Two sets have equivalent cardinality if, and only if, there exists a bijection between them. Thus, given the information $M,N$ are sets with $f$ a bijection between them we can really only deduce $M,N$ have the same cardinality (under some very technical assumptions if I remember correctly). For this reason, we would say $M,N$ are isomorphic as sets with $f$ a set isomorphism between $M$ and $N$.
Now let's take a look at something more general. Let's say $M$ is a set and $\tau$ is a structure on $M$. We can write this as $(M,\tau)$ if we would like to. Now suppose we have another set $N$ with a structure (of similar "type") called $\sigma$. Again, we can keep track of this set and its structure by the ordered pair $(N,\sigma)$. As before, if we are given a bijection, $$M\overset{f}\longrightarrow N$$ we can really only conclude $M,N$ are isomorphic as sets. However, let's say that $\tau$ gives $M$ a vector space structure (i.e. elements of $M$ can be added and scalar multiplication from a field $K$ is allowed). This is where the same "type" is important. If $\sigma$ is a topology on $N$, there isn't an immediate (to me it is not immediate) way to show $\tau$ and $\sigma$ are pretty much the same. So assume $\sigma$ is also a vector space structure but on $N$. We want to describe an isomorphism between $M$ and $N$ so that the only real difference between $(M,\tau)$ and $(N,\sigma)$ is how we labeled them. One way to accomplish this is to observe whenever $f(ka+b)=kf(a)+f(b)$ for $a,b\in M$ and $k\in K$ it does not matter if we compute the operations of $+, \cdot$ in $M$ first or in $N$, after we map with $f$. Since $f$ was assumed to be a bijection, we have that $M,N$ are the same size as sets (same cardinality), and all of the operations are the same. So $(M,\tau)$ and $(N,\sigma)$ are isomorphic as vector spaces and $f$ is a vector space isomorphism between them.
We can extend this idea above to any finite number of structures on a set $M$ - if we have $\tau_1, \tau_2,\dots, \tau_n$ we look at $(M,\tau_1,\dots, \tau_n)$ - but now this comes down to our idea of what a structure is. There probably isn't a definitive definition for this yet (maybe there is, but I would wager in a lot of cases this definition is from intuition built on many examples and uses). So let's say
Definition: A structure on a set $M$ is any extra information we are given about the behavior of elements, functions, or subsets of $M$. This captures the idea of an algebraic structure (a set $M$ with a binary operation $\phi$), a topological structure (a set $M$ with a collection of open subsets $\tau$), an ordering (a set $M$ with a relation $\preceq$), and many other ideas. We say that $(M,\tau_1,...,\tau_n)$ is an object $M$ with the structures $\tau_1,\ldots,\tau_n$.
This also motivates the
Definition: Two objects with given structure $(M,\tau)$ and $(N,\sigma)$ are said to be isomorphic under the given "type" of structure if $(M,\tau)$ and $(N,\sigma)$ are the same up to relabeling of elements.
Mathematically, these are pretty informal definitions. But they can be given rigor through set theory.
Edit: Let's show that there is some truth in these definitions (with or without justification).
We will say a structure is valid or consistent if it does not give rise to any illogical propositions. For the rest of this, all structures are assumed valid.
An object $(M,\tau_1,...,\tau_n)$ is a set $M$ with any finite number of structures $\tau_1,...,\tau_n$. We call $(U, \tau_1|_U,...,\tau_n|_U)$ a subobject if $U\subseteq M$ and $\tau_1|_U,...,\tau_n|_U$ are the induced structures on this restriction.
Given a map $f:M\rightarrow N$ of sets with associated structures $\tau$ and $\sigma$ respectively, we will call $f$ a morphism if $f$ induces a valid substructure in $\sigma$. More formally, if we denote by $\hat{f}(\tau)$ the substructure induced by $f$, then $(f(M),\hat{f}(\tau))$ is a subobject of $N$.
Proposition: If $(M,\tau_1,...,\tau_n)$, $(N,\sigma_1,...,\sigma_m)$ are objects and $f$ is an injective morphism (where an injective morphism means a morphism whose mapping is an injection on the level of sets) from $M$ to $N$, then $M$ is isomorphic to a subobject of $N$.
Proof. Since $f$ is a morphism, $(f(M),\hat{f}(\tau_1),...,\hat{f}(\tau_n))$ is a subjobect of $N$ (if $n\geq m$ it may be the case that $f$ "forgets" some of the structure). Since $f$ is an injection, for any $x\in M$ we have a distinct $y=f(x)\in N$. So we may relabel $x$ as $y$. Now we just need to check that structure is preserved. But this is true since $\hat{f}(\tau_i)=\sigma_j|_{f(M)}$ for some $i,j$ by definition.$\square$
In fact, this is a generalization of the first isomorphism theorem when $f$ is an injection.
