Why can't the density difference between the liquid and solid be an appropriate order parameter for liquid-to-solid transition?

The reason why the density difference is not used as an order parameter for a liquid to solid transition is that the density does NOT actually describe the difference between a liquid and a solid!

In fact, you could even have a phase transition between a liquid and solid that have exact the same mass density! In some cases, like for water, the solid is less dense than the liquid phase. Density is not the main difference between the two phases.

So what is the actual difference between a solid and liquid?

A solid is characterized by static correlations of atomic positions, while in a liquid the atomic positions are uncorrelated for long-enough time scales. In other words, if you see two atoms in a liquid and wait long enough their positions will change with respect to each other. However, if you do the same in a solid, their positions will stay fixed with respect to each other for basically forever.

Like the other answer states, it is possible to also have additional crystalline order, but that is very rarely the case (normally it is to an amorphous solid). I will discuss the general case below.

The order parameter for the liquid-solid transition would involve something like the atomic position correlation function $G(x,x^{\prime};t-t^{\prime})$, which is basically the probability you will sight an atom in position $x^{\prime}$ at time $t^{\prime}$ if you see an atom at $x$ at time $t$.

For a liquid, the following limit is obeyed for $G(x,x^{\prime};t-t^{\prime})$:

$$\lim_{t\to\infty} G(x,x^{\prime};t-t^{\prime})=0$$

That is to say, there is no correlation between seeing two atoms over long time scales. This makes sense because if you look at two atoms in a liquid and wait long enough, both atoms will have moved to new positions in a random fashion.

However for a solid, the atomic positions are highly correlated. If you see two atoms next to each other today, they will be in the same places tomorrow, the day after, next year, etc. etc. This means that:

$$\lim_{t\to\infty} G(x,x^{\prime};t-t^{\prime}) \neq 0$$

Now that we have a quantity that is zero in one phase, and non-zero in another phase, we have defined the order parameter $\langle \tilde \rho \rangle$:

$$\langle \tilde \rho(x,x^{\prime}) \rangle \equiv \lim_{t\to\infty} G(x,x^{\prime};t-t^{\prime}) \equiv \tilde G(x,x^{\prime})$$

For a crystal, $\langle \tilde \rho(x,x^{\prime}) \rangle$ will have special periodic structure in $x,x^{\prime}$, but in general that isn't the case.