My answer. The importance of unitary operators in QM relies upon a pair of fundamentals theorems, known as Wigner's and Kadison's theorem respectively.

Consider a quantum system described in a Hilbert space ${\cal H}$. Pure states are represented by vectors $\psi \in \cal H$ with $||\psi||=1$. Actually a state is the whole set $[\psi ] := \{e^{ia} \psi \:|\: a \in R\}$ when $||\psi||=1$, because there is no way to fix the phase $e^{ia}$ and all physical quantities are independent form it. Notice that, with the introduced notation, one hence has $[\psi]= [\phi]$ if and only if $\psi = e^{ia} \phi$ for some $a \in R$. Henceforth $S(\cal H)$ will denote the space of pure states in $\cal H$. (It coincides to the complex projective space of $\cal H$ but it does not matter now.)

If you have two states $[\psi] \neq [\phi]$, represented by normalized vectors $\psi$ and $\phi$, as is well known $$P([\psi],[\phi])| := |\langle \phi|\psi \rangle|^2$$ is a physical quantity representing the probability transition from the former to the latter state (in both directions). Notice that, as it mast be, that quantity is not affected by changes of the arbitrary phases embodied in the states.

(1W) the map is bijective ("into" and "onto"). In other words, with a given symmetry one can move form a given state to any other state, and the action of the symmetry is reversible;

There are lots of physical transformations of states verifying (1W) and (2W). For instance: time evolution, charge conjugation, parity reflection, time reversal, the action of Poincaré's group o Galileo's group, and many many others. Sometimes it turns out that a supposed symmetry does not exists (like parity reflection for systems subjected to weak interaction). The natural question is therefore:

In particular, taking Wigner's theorem into account we may forget of states as class of vectors and we can safely use vectors as usual to represent states. (Life is not so easy, since an annoying and difficult problem arises as soon as one try to represent groups of symmetries this way)

Let us come to the other important theorem due to Kadison. Everybody knows that there exists another, more general, notion of state in QM. I mean a mixed state. A mixed state is represented by a positive, trace class operator $\rho : \cal H \to \cal H$ with unit trace $tr(\rho)=1$. The set $M(\cal H)$ of mixed states is a convex body in the real linear space of self-adjoint bounded operators on $\cal H$. This means that if $p_1,\ldots, p_n \in [0,1]$ and $\rho_1,\ldots, \rho_n \in M(\cal H)$, then $$\sum_{i=1}^n p_i \rho_i \in M({\cal H})\:. \qquad (1)$$

Pure states $[\psi]$ are a subcase of mixed ones, those of the form $|\psi\rangle \langle \psi|$ verify $\sum_i p_i =1$ (notice that the arbitrary phase affecting $\psi$ does not matter here). Actually it is possible to prove that pure states are all of the states in $M(\cal H)$ such that cannot be decomposed as in (1) (for some non-trivial decomposition): They are the extremal elements of $M(\cal H)$.

Every mixed state $\rho$, in general, admits many decompositions like that in (1). One is the most natural, that obtained by the spectral decomposition: $$\rho = \sum_{k=0}^{+\infty} p_k |\psi_k \rangle \langle \psi_k|\:,\qquad (2)$$ where now the $p_k$s are the eigenvalues of $\rho$, that (by hypotheses on $\rho$) belong to $[0,1]$ and their sum is $1$. (2) has a natural interpretation as a classical mixing, with classical probabilities $p_k$, of pure quantum states $|\psi_k \rangle \langle \psi_k|$. Actually in general there are many ways to decompose $\rho$ into such a form, it is untenable separating classical probability (carried by the $p_n$s) and quantum probability (embodied in the $|\psi_k \rangle \langle \psi_k|$s).

Within this context, there is another notion of quantum symmetry due to Kadison (actually it was formulated using a dual approach and referring to the lattice of projectors).

This is a notion of quantum symmetry seems to be quite different from that proposed by Wigner. Nonetheless the two notions coincide in view of the celebrated (however less known than Wigner's one) Kadison's theorem.

My answer. The importance of unitary operators in QM relies upon a pair of fundamental theorems, known as Wigner's and Kadison's theorem respectively.

Consider a quantum system described in a Hilbert space ${\cal H}$. Pure states are represented by vectors $\psi \in \cal H$ with $||\psi||=1$. Actually a state is the whole set $[\psi ] := \{e^{ia} \psi \:|\: a \in R\}$ when $||\psi||=1$, because there is no physical way to fix the phase $e^{ia}$ and all physical quantities are independent form it. Notice that, with the introduced notation, one hence has $[\psi]= [\phi]$ if and only if $\psi = e^{ia} \phi$ for some $a \in R$. Henceforth $S(\cal H)$ will denote the space of pure states in $\cal H$. (It coincides to the complex projective space of $\cal H$ but it does not matter now.)

If you have two states $[\psi] \neq [\phi]$, represented by normalized vectors $\psi$ and $\phi$, as is well known $$P([\psi],[\phi])| := |\langle \phi|\psi \rangle|^2$$ is a physical quantity representing the probability transition from the former to the latter state (in both directions). Notice that, as it must be, that quantity is not affected by changes of the arbitrary phases embodied in the states.

(1W) the map is bijective ("into" and "onto"). In other words, with a given symmetry one can moves form a given state to any other state, and the action of the symmetry is reversible;

There are lots of physical transformations of states verifying (1W) and (2W). For instance: time evolution, charge conjugation, parity reflection, time reversal, the action of Poincaré's group or Galileo's group, and many many others. Sometimes it turns out that a supposed symmetry actually does not exist (like parity reflection for systems subjected to weak interaction). 

The natural question is therefore:

In particular, taking Wigner's theorem into account, we may forget of states as classes of vectors and we can safely use vectors as usual to represent states. (Life is not so easy, since an annoying and difficult problem arises as soon as one tries to represent a group of symmetries this way)

Let us come to the other, analogous, important theorem due to Kadison. Everybody knows that there exists another, more general, notion of state in QM. I mean a mixed state. A mixed state is represented by a positive, trace class operator $\rho : \cal H \to \cal H$ with unit trace $tr(\rho)=1$. The set $M(\cal H)$ of mixed states is a convex body in the real linear space of self-adjoint bounded operators on $\cal H$. This means that if $p_1,\ldots, p_n \in [0,1]$ and $\rho_1,\ldots, \rho_n \in M(\cal H)$ verify $\sum_i p_i =1$, then: $$\sum_{i=1}^n p_i \rho_i \in M({\cal H})\:. \qquad (1)$$

Pure states $[\psi]$ are a subcase of mixed ones, those of the form $|\psi\rangle \langle \psi|$ (notice that the arbitrary phase affecting $\psi$ does not matter here). Actually it is possible to prove that pure states are all of the states in $M(\cal H)$ such that cannot be decomposed as in (1) (for some non-trivial decomposition): They are the extremal elements of $M(\cal H)$.

Every mixed state $\rho$, in general, admits many decompositions like that in (1). One is the most natural, that obtained by the spectral decomposition: $$\rho = \sum_{k=0}^{+\infty} p_k |\psi_k \rangle \langle \psi_k|\:,\qquad (2)$$ where now the $p_k$s are the eigenvalues of $\rho$, that (by hypotheses on $\rho$) belong to $[0,1]$ and their sum is $1$. (2) has a natural interpretation as a classical mixing, with classical probabilities $p_k$, of pure quantum states $|\psi_k \rangle \langle \psi_k|$. Actually in general there are many ways to decompose $\rho$ into such a form, it is untenable separating classical probabilities (carried by the $p_n$s) and quantum probabilities (embodied in the $|\psi_k \rangle \langle \psi_k|$s).

Within this context, there is another notion of quantum symmetry due to Kadison (actually it was formulated using a dual, equivalent, approach referring to the lattice of orthogonal projectors).

Such a notion of quantum symmetry seems to be quite different from that proposed by Wigner. Nonetheless the two notions coincide in view of the celebrated (however less known than Wigner's one) Kadison's theorem.

The importance of unitary operators in QM relies upon a pair of fundamentals theorems, known as Wigner's and Kadison's theorem respectively.

ADDED REMARK: This is an answer to Antillar Maximus' question:

"A more general question would be, why is a unitary transformation useful?"

My answer. The importance of unitary operators in QM relies upon a pair of fundamentals theorems, known as Wigner's and Kadison's theorem respectively.