No, entangled particles need not be indistinguishable.
Consider two particles $A$ and $B$, which correspond to single-particle Hilbert spaces $\mathcal H_A$ and $\mathcal H_B$. The Hilbert space underlying the composite system is the tensor product space $\mathcal H = \mathcal H_A \otimes \mathcal H_B$.
Question: Can all states in $\mathcal H$ be written in the form $\alpha \otimes \psi$, where $\alpha \in \mathcal H_A$ and $\psi \in \mathcal H_B$? The answer is no - while this may be possible for some states, in general elements of $\mathcal H$ cannot be written in this way.
Consider for example the state
$$\Psi = \alpha \otimes \psi + \beta \otimes \psi + \alpha \otimes \phi + \beta \otimes \phi$$
where $\alpha,\beta \in \mathcal H_A$ and $\psi,\phi\in\mathcal H_B$. It isn't difficult to see that this state can be "factored"
$$\Psi= (\alpha + \beta) \otimes (\psi + \phi)$$
and therefore falls into the above category. However, states such as
$$\Phi = \alpha \otimes \psi + \beta \otimes \phi$$
cannot be written in this way. States such as $\Phi$ which cannot be factored into a single tensor product between an element of $\mathcal H_A$ and an element of $\mathcal H_B$ are called entangled.
Nowhere in the above description of entanglement did I imply or require that $\mathcal H_A$ and $\mathcal H_B$ correspond to identical particles. They need not even be copies of the same Hilbert space - we could have $\mathcal H_A = L^2(\mathbb R^3)\otimes \mathbb C^2$ (corresponding to a spin-1/2 particle) and $\mathcal H_B = L^2(\mathbb R^3)$ (corresponding to a spin-0 particle). But even if $\mathcal H_A = \mathcal H_B$, the particles need not be indistinguishable.
That two particles are indistinguishable means (1) that $\mathcal H_A = \mathcal H_B$, and (2) that every state $\Psi \in \mathcal H$ is mapped to $\Psi' = e^{i\theta} \Psi$ under particle interchange (in which $\alpha\otimes\psi \mapsto \psi \otimes \alpha$). In most cases, $\theta= \{0,\pi\}$ corresponding to bosons and fermions, respectively (though this does not always hold in 2D, where we can have more exotic possibilities).
In any case, this added constraint of indistinguishability is downstream of the notion of entanglement, which requires only that the Hilbert space be written as a tensor product of simpler Hilbert spaces.
Have you any idea on what John Rennie might have meant by the two particles become mixed up?
The two particles become "mixed up" in the sense that it is no longer meaningful to speak about the state of either particle independently of the other.
Consider two (not necessarily indistinguishable) non-interacting particles in a box of length $L$. One possible state for this composite system is
$$\Psi = \left(\frac{\psi_1+\psi_2}{\sqrt{2}}\right) \otimes \left(\frac{\psi_3 + 2\psi_4}{\sqrt{5}}\right)$$
where $\psi_n$ is the $n^{th}$ energy eigenstate for the single particle in a box.
In this state, measurements of the energies of the two particles are independent in the sense of statistical independence. Physically, this means that a measurement of the first particle gives me no information whatsoever about the second particle; in this sense, we can think of them as two separate particles (upon which we can perform two separate measurements) going about their business.
On the other hand, consider an entangled state
$$\Phi = \frac{1}{\sqrt{5}} \left[ \psi_1 \otimes \psi_3 + 2\psi_2 \otimes \psi_4 \right]$$
In this state, the measurements are no longer independent. A measurement of the state of the first particle constitutes a measurement of the state of the second as well; if I find the first particle to have energy $E_1$, then I will find the energy of the second particle to be $E_3$ with 100% probability.
In this sense, it is no longer meaningful to talk about the state of one particle or the other. We can only talk about the state of the system of two particles.
Incidentally, this is the origin of much confusion among laymen and non-laymen alike. People intuitively feel that if you measure the first particle by itself, then some superluminal influence must propagate out to the second, but this is wrong. In truth, there is no such thing as a measurement of the first particle alone.
