This question raises several issues simultaneously requiring a closer scrutiny before answering the question itself.
- Complex conjugation.
If ${\cal H}$ is a complex Hilbert space and $B=\{u_i\}_{i\in I}$ is a Hilbert basis therein, an antilinear operation is defined
$$K_B : \sum_i c_i u_i \mapsto \sum_i c^*_i u_i\tag{1}$$
called complex conjugation associated to $B$.
Antilinearity means $$K_B(au+bv) = a^*K_Bu + b^*K_Bv\quad \forall u,v \in {\cal H} \quad \forall a,b \in \mathbb C.$$
It is evident that this definition depends on the choice of $B$. If $B'= \{u'_j\}_{j\in I}$ is another Hilbert basis of ${\cal H}$, it turns out that
$$K_B=K_{B'} \quad \Leftrightarrow \quad \langle u_i|u'_j\rangle = \langle u_i|u'_j\rangle^* \quad \forall i,j \in I\:.$$
From the definition, we have that $$K_BK_B =I\tag{2}$$ and $$\langle K_B v|K_B u\rangle =
\langle v|u\rangle^* \quad \forall u,v \in {\cal H}\:,\tag{3}$$
so that, in particular, $K_B$ is isometric because
$$||K_Bu||= ||u||\quad \forall u \in {\cal H}\:.$$
As an example, consider ${\cal H}= L^2(\mathbb R, dx)$, the Hilbert space of a
spinless particle living along the real line.
A natural conjugation is defined like this.
$$K : L^2(\mathbb R, dx) \ni \psi \mapsto \psi^*\:,$$
where
$$\psi^*(x) := (\psi(x))^*\quad \forall x \in \mathbb R\:.$$
It is quite simple to prove that
$$K=K_B\:,$$
where $B=\{f_n\}_{n \in \mathbb N}$ is the Hilbert basis
$$f_n(x) = \frac{H_n(x)e^{-x^2/2}}{\sqrt{\pi^{1/2} 2^n n!}}$$
constructed out of real Hermite polynomials $H_n$, defining in particular an orthonormal basis of eigenstates of the Hamiltonian of the harmonic oscillator usually denoted by $f_n = |n\rangle$. Notice that no complex phases are permitted.
Another possible conjugation is $K'$ obtained with the same recipe as $K$, but working in momentum picture,
$$K' : \psi \mapsto F^{-1} ((F \psi)^*)\:,$$
where
$$F(\psi)(p) := \hat{\psi}(p) := \frac{1}{\sqrt{2\pi}} \int e^{-ipx} \psi(x) dx$$
is the standard Fourier-Plancherel transform of $\psi$ (I assume $\hbar=1$).
It is easy to check that $$K\neq K'\:.$$
- Notion of complex conjugation of an operator
Given a linear operator $A : D(A) \to {\cal H}$, where the domain $D(A)\subset {\cal H}$ is a subspace, and a conjugation $K_B : {\cal H} \to {\cal H}$ (depending on the basis $B$), it is possible to define another linear operator $A^{*_{K_B}}$ that we may call the complex conjugated operator of $A$ with respect to $K_B$.
$$A^{*_{K_B}} := K_B A K_B\tag{*}$$
provided $K_B (D(A)) \subset D(A)$. I stress that $K_B$ appears twice in the right-hand side of the definition above. This is because we want that $A^{*_B}$ is linear as $A$ is:
A definition like this
$$A^{*_{K_B}} := K_B A\quad \mbox{(wrong),}$$
would instead produce an antilinear operator:
$$(K_BA)(au) = K_B(aA(u))= a^* K_BAu\:.$$
Also observe that, in the absence of issues with the domains of the operators, (*) implies $$(AB)^{*_{K_B}} = A^{*_{K_B}}B^{*_{K_B}}\tag{4}\:.$$
Consider for instance the momentum operator restricted to the subspace of Schwartz' functions ${\cal S}(\mathbb R)$. As is well known,
$$P\psi = -i \frac{d}{dx} \psi \:,\quad \psi \in {\cal S}(\mathbb R)\:.$$
It is immediately proved that, referring to conjugations $K$ and $K'$ discussed in 1,
$$P^{*_K} = +i \frac{d}{dx} = -P\:,\tag{5}$$
whereas
$$P^{*_{K'}} = -i \frac{d}{dx} = P\:.$$
- Antiunitary operators.
Consider an antiunitary operator $V : {\cal H} \to {\cal H}$. By definition it is antilinear and $$\langle V u|V u\rangle =
\langle u|v\rangle^* \quad \forall u,v \in {\cal H}\:. \tag{3'}$$
If $B$ is a Hilbert basis and $K_B$ the associated conjugation, we have from (2),
$$V = VK_BK_B$$
so that $U_B := VK_B$ is linear and $V_B$ is also unitary
$$\langle U_B u|U_B v\rangle =
\langle u|v\rangle \quad \forall u,v \in {\cal H}\:,$$
from (3) and (3').
We conclude that
Proposition. Given a Hilbert basis $B$, an antiunitary operator $V$ can be decomposed as $V= U_B K_B$ where $U_B$ is unitary and $K_B$ is the conjugation associated to the said basis.
- OP's question.
The question have an overall elementary answer without entering into the details of unitary/antiunitary operators. In fact, both unitary or antiunitary operators are functions $f: {\cal H} \to {\cal H}$ and the composition of functions (linear or antilinear operators does not matter) is associatve.
OP's mistake relies on the identity $\Theta A =A^*$ that has no clear meaning.
Let us consider the simplest case of $\Theta = K_B$.
(i) Suppose that $K_B A =A^*$ is the definition of the right-hand side.
This has nothing to do with the correct definition of conjugated operator with respect to a basis as I defined in 2 because this $A^*$ is antilinear, whereas the complex conjugated of an operator with respect to a basis is linear. Also it does not produce natural identities like (5)
$$\left(-i\frac{d}{dx}\right)^* = i\frac{d}{dx}$$
established above, where the relevant conjugation is the standard conjugation of wavefunctions (I denoted by $K$). Finally, this definition of $A^*$ would not satisfy (4) and the contradiction suggested by the author is based on the validity of (4).
(ii) Suppose that $K_B A =A^*$ is not the definition of the right-hand side. In this case, author's argument immediately stops since we do not know what the right-hand side is.
With the right definition of conjugated operator (with respect to a Hilbert basis) as in (*) everything goes right also because $K_B A \neq A^{*_{K_B}}$.
