A discussion occurred between us in chat, so here I will highlight the end results.
Depending on how exactly you define a tuple, there are a number of options, but the end result is that the operation is essentially the set of all elements appearing in at least one of $t_1$ or $t_2$ placed into a tuple with entries in the same order as they would appear in the underlying set.
Interpreting tuples as ordered sets:
Letting $t_1 = (A,\prec_a)$ and $t_2 = (B,\prec_b)$ and the underlying set be $(X,\prec_x)$ of which both $A$ and $B$ are subsets of $X$, we have:
$$t_1*t_2 = \left((A\cup B), \prec_x\cap (A\cup B)^2\right)$$
keeping in mind that an order on $X$ is itself a set and is a subset of the cartesian product of $X$ with itself. The intersection them makes sense. Alternatively, one could just refer to the resulting order for $t_1*t_2$ as the inherited order, for which various notations exist such as $\prec_x|_{A\cup B}$.
Interpreting tuples as injective functions from $[n]$ to $X$:
Letting $(X,\prec)$ be the underlying ordered set and letting $t_1=f_a$ where $f_a\in X^{\underline{[n_a]}}$ (i.e. letting $f_a$ be an injective function from $[n_a]$ to $X$) and $t_2=f_b\in X^{\underline{[n_b]}}$, we have:
$$t_1*t_2 = g\in (f_a(A)\cup f_b(B))^{[|f_a(A)\cup f_b(B)|]}~\text{satisfying}~\\\forall j,k\in[|f_a(A)\cup f_b(B)|]~\text{one has}~j<k\implies g(j)\prec g(k)$$
Both of these seem like a lot of words to use for something that can be described much more simply with english rather than symbols: $t_1*t_2$ is the unique tuple such that each entry of $t_1$ and $t_2$ appear once and they appear in the same order as the underlying set.
Best Answer
The explanation identifies a $1$-tuple with its object. So while $x$ is generally not the same as $\{x\}$, the $1$-tuple $(x)$ is identified with $x$ itself.
This is perfectly reasonable to do, and so people do it. But does that clash with the set theoretic definition? Well, that depends on your definition of a $1$-tuple to begin with. Since the language of set theory only has $\in$, any definition of $1$-tuples is just a way to interpret this concept of a sequence of length $1$.
Generally, we think about $n$-tuples as functions from $\{0,\ldots,n-1\}$, so a $1$-tuple would be a function from $\{0\}$ into something. But this is not something which is hardcoded into set theory, it's just one way to interpret $n$-tuples. You can do it just as well in a different way, or even explicitly decide this is fine for $n>1$, and for $n=1$, $(x)$ is just $x$ itself (which then raises the question what is a $1$-tuple whose object is a $3$-tuple?).
This is also why we can't require $\{x\}=x$ in general. Since we understand $\{x\}$ as a set whose unique member is $x$. And then $\varnothing\neq\{\varnothing\}$, since exactly one of these has an element. But tuples are looser, since as we see, they are not hardcoded into set theory like singletons.