It is indeed always a semiprime. Think about it this way:
Consider a number $x = qq_1$ for some primes $q$ and $q_1$. We know that if $p$ is a prime and $p|ab$, then $p|a$ or $p|b$. Well, what if some other prime $p|x$? This would imply that $p|q$ or $p|q_1$, which is not possible because $q$ and $q_1$ are themselves prime. Therefore, the only possible nontrivial divisors of $x$ are $q$ and $q_1$.
The definition of irreducibles which is mentioned in the question is perfect.
Now you may observe that there are elements (say $p$ ) in $T$ such that for any $a,b \in T$, $p$ divides $ab$ implies that $p$ divides $a$ or $b$. We call such elements as prime in $T$. For example, $7$ is a prime in $T$ because $7$ is a prime in $\mathbb{N}$ and we can make use of Euclid's lemma to conclude. Note that $7$ is also irreducible.
Also we can see that $10$ is not a prime in $T$. To see that we take the same example which you provided, take $ab = 100, a= 4, b =25$, so $10$ divides $100$ but none of $a,b$.
It is possible that an irreducible $p$ (such as $10$) can divide $1^{e_1}\cdot 4^{e_4}\cdot 7^{e_7}\cdot 10^{e_{10}}... $ without dividing any of $1,4,7, \dots$, etc.This was a flaw in the argument.
You can easily apply your argument to the sets where each irreducible is a prime in that set.
Update:
I think OP considering as Euclid's lemma is the generalization of Euclid's lemma given here - https://en.wikipedia.org/wiki/Euclid%27s_lemma .
It says that
Euclid's lemma — If a prime $p$ divides the product $ab$ of two integers $a$ and $b$, then $p$ must divide at least one of those integers $a$ and $b$.
Note that the lemma does not make use of the concept of $\gcd$.
I don't think we can show that the lemma does not hold in $T$ without exhibiting counterexamples. When we define a new structure on sets, we appropriately define a set of rules or axioms which the elements of the set should satisfy.
Best Answer
The set of integers is Unique Factorization Domain, that is, every element can be uniquely factorized. In UFD, a non-unit element is prime if and only if it is irreducible. So in $\mathbb{Z}$, there is no counterexample.
In $\mathbb{Z}[\sqrt{5}]$, which is a set of elements of the form $a+b\sqrt{-5}$ where $a$ and $b$ are integers, $9$ can be written in two forms, $$ 9=3^2 = (2-\sqrt{-5})(2+\sqrt{-5}) $$ and $3$ divides $(2-\sqrt{-5})(2+\sqrt{-5})$ but does not divide neither $2-\sqrt{-5}$ nor $2+\sqrt{-5}$. So $3$ is not a prime here. You can also see that $\mathbb{Z}[\sqrt{-5}]$ is not UFD. However, it is an irreducible element. This can be shown by solving the following equation $$ 3=(a+b\sqrt{-5})(c+d\sqrt{-5}) $$ for integers $a,b,c$ and $d$. Long and uninteresting calculation will show that $b=d=0$ and either one of $a$ and $c$ is $1$, and the other one is $3$.
Conversely, in integral domain $R$, being prime always implies that it is an irreducible element. Suppose a prime $p$ is reducible, so there are two non-unit elements $a,b$ such that $p=ab$. Since $p$ divides $ab=p$, by the definition of the primality, $p$ must divide $a$ or $b$. We may assume that $p$ divides $a$, and there is $c\in R$ such that $pc=a$. Thus $$ p=ab=pcb \implies cb=1 $$ so it contradicts that $b$ is non-unit. The cancellation works because we are working in the integral domain.