I think you have all the knowledge you need. The definitions are as you state, and Unitarity implies reversibility (all unitary transformations have an inverse) but the converse doesn't hold. To prove that they are not the same concept, you simply need to show that there is a reversible, but non unitary transformation. Enter exhibit A: the following linear homogeneous transformation:

$$\mathscr{L}:(\mathbb{C}^2,\,+)\to (\mathbb{C}^2,\,+);\;\;\mathscr{L}\left(\begin{array}{c}z_1\\z_2\end{array}\right) = \left(\begin{array}{cc}1&z\\0&1\end{array}\right)\left(\begin{array}{c}z_1\\z_2\end{array}\right)$$

All matrix groups comprise only invertible transformations. Most of them are not unitary transformations: the following is a strict set inclusion $SU(2)\subset SL(2,\,\mathbb{C})$.

Is this what you mean? (I'm having a little trouble gauging your level, as I have seen some good answers from you, so I am a little surprised at your question - not being critical - I lack all kinds of knowledge that I should have too).

Now if your question is specifically to do with quantum mechanics, then quantum states always need to have an $\mathcal{L^2}$ length of one, if you accept the Born interpretation of the quantum state as a probability amplitude vector. All possibilities must sum to unity probability: the system must be in some state and, to be a full specification of the quantum state, the state vector must exhaust possibilities. So any realistic transformation of a quantum state must preserve the length. Any reversible linear transformation between quantum states must be unitary. Indeed, one doesn't have to assume linearity if one assumes that transformations wrought on quantum states are compatible with Lorentz transformations on the physical system, then the transformation must be linear and unitary; this is part of what Wigner's theorem tells us (which I talk about more in my answer here).

I think you have all the knowledge you need. The definitions are as you state, and Unitarity implies reversibility (all unitary transformations have an inverse) but the converse doesn't hold. To prove that they are not the same concept, you simply need to show that there is a reversible, but non unitary transformation. Enter exhibit A: the following linear homogeneous transformation:

$$\mathscr{L}:(\mathbb{C}^2,\,+)\to (\mathbb{C}^2,\,+);\;\;\mathscr{L}\left(\begin{array}{c}z_1\\z_2\end{array}\right) = \left(\begin{array}{cc}1&z\\0&1\end{array}\right)\left(\begin{array}{c}z_1\\z_2\end{array}\right)$$

All matrix groups comprise only invertible transformations. Most of them are not unitary transformations: the following is a strict set inclusion $SU(2)\subset SL(2,\,\mathbb{C})$.

Is this what you mean? (I'm having a little trouble gauging your level, as I have seen some good answers from you, so I am a little surprised at your question - not being critical - I lack all kinds of knowledge that I should have too).

Now if your question is specifically to do with quantum mechanics, then quantum states always need to have an $\mathcal{L^2}$ length of one, if you accept the Born interpretation of the quantum state as a probability amplitude vector. All possibilities must sum to unity probability: the system must be in some state and, to be a full specification of the quantum state, the state vector must exhaust possibilities. So any realistic transformation of a quantum state must preserve the length. Any reversible linear transformation between quantum states must be unitary. Indeed, one doesn't have to assume linearity if one assumes that transformations wrought on quantum states are compatible with Lorentz transformations on the physical system, then the transformation must be linear and unitary; this is part of what Wigner's theorem tells us (which I talk about more in my answer here).