Short answer: During the collision, while the first ball is going faster than the second, the balls are compressing. Once the second ball begins moving as fast as first, the balls start uncompressing, pushing the balls apart. Once they fully separate, they are moving different speeds.
Medium answer:
The first half of the collision pushes the billards together and deforms them, and in the second half of the collision, the energy stored in the deformation is released back into mechanical energy, and the billards undeform and push apart, resulting in different velocities. Billards are "elastic", or like springs: almost all of the energy when deforming them is stored and mechanically released.
This elastic deformation is the same reason when you throw a billard against a wall, it doesn't stick to the wall; it bounces back. Inelastic materials, like a ball of Play-Doh, will stick to the wall. 621
(Or other balls of Play-Doh.)
Long answer: In addition, the earlier answers, you can reach the result mathematically with physical laws and one assumption.
A result of Newton's Laws is the conservation of momentum
$$m_a\mathbf{v_{a1}} + m_b\mathbf{v_{b1}}=m_a\mathbf{v_{a2}}+m_b\mathbf{v_{b2}}$$
In this case, $\mathbf{v_{b1}}=\mathbf{0}$ and $m_a=m_b$, so
$$\mathbf{v_{a1}}=\mathbf{v_{a2}}+\mathbf{v_{b2}}$$
Obviously, there are many values that satisfy this equation. So we need add another physical law: the Law of Conservation of Energy. (Assumption: there is negligible rotation of the balls.)
$$\frac{m_a\mathbf{v_{a1}}^2}{2}+\frac{m_b\mathbf{v_{a1}}^2}{2}=\frac{m_a\mathbf{v_{a2}}^2}{2}+\frac{m_b\mathbf{v_{b2}}^2}{2} + \Delta H$$ $$\mathbf{v_{a1}}^2=\mathbf{v_{a2}}^2+\mathbf{v_{b2}}^2 + \frac{2\Delta H}m$$
Because billards are elastic, none of the energy is converted into heat (this in fact, is what is meant by elastic...you could go further into materials science and find out exactly what about the chemistry of the billards causes this property, but at the end of the day, billards are elastic, like bouncy balls or springs), and
$$\mathbf{v_{a1}}^2=\mathbf{v_{a2}}^2+\mathbf{v_{b2}}^2$$ $$\mathbf{v_{a1}}^2=(\mathbf{v_{a1}} - \mathbf{v_{b2}})^2+\mathbf{v_{b2}}^2$$ $$\mathbf{v_{a1}}^2=\mathbf{v_{a1}}^2 - 2\mathbf{v_{a1}}\mathbf{v_{b2}} + \mathbf{v_{b2}}^2 + \mathbf{v_{b2}}^2$$ $$\mathbf{v_{a1}}\mathbf{v_{b2}} = \mathbf{v_{b2}}^2$$
For a straight-on collision, $\mathbf{v_{a1}}$ and $\mathbf{v_{b2}}$ are parallel, so
$$|\mathbf{v_{a1}}||\mathbf{v_{b2}}| = |\mathbf{v_{b2}}|^2$$
Since billards cannot pass through one another $\mathbf{v_{b2}} \neq 0$, and thus $\mathbf{v_{b2}} = \mathbf{v_{a1}}$.
Note that this result depends on the assumption $\Delta H = 0$; i.e. the balls un-deformation converted 100% of the energy from the deformation back into mechanical energy. I.e. an elastic collision., 512