Momentum conservation states that:

$$m_{1} \mathbf v_{1i} + m_{2} \mathbf v_{2i}=m_{1} \mathbf v_{1f} + m_{2} \mathbf v_{2f}$$

where $m_1$ and $m_2$ are the masses of the objects, $\mathbf v_{1i}$ and $\mathbf v_{2i}$ are the initial velocities and $\mathbf v_{1f}$ and $\mathbf v_{2f}$ are the initial velocities. In your specific cas the momentum equation reduces to

\begin{align} m\mathbf v + m (-\mathbf v) & = m\mathbf v _1 + m\mathbf v _2\\ 0&=m\mathbf v _1 +m \mathbf v _2\\ 0&=\mathbf v _1 + \mathbf v _2 \tag{1} \end{align}

where $m$ is the mass of the objects, $\mathbf v$ is the initial velocity of both the objects and $\mathbf v_1$ and $\mathbf v_2$ are the final velocities.

Now there are two variables of interest ($\mathbf v_1$ and $\mathbf v_2$) but there is only one equation. So as you can see, momentum conservation cannot be used alone to predict the final velocities of two colliding objects. You need to apply some other constraint/equation which will then help you in uniquely determining the final velocities.

In this case, since you are talking about elastic collisions, we can apply two equivalent constraints, one being energy conservation other being $e=1$ (where $e$ is the coefficient of restitution).

Note that from $(1)$, $|\mathbf v_1| = |\mathbf v_2 |$. From now on, I will denote the magnitudes of velocities as $v_1$, $v_2$ and $v$. Thus applying energy conservation, we get

\begin{align} \frac 1 2 m v^2+ \frac 1 2 m v^2 &= \frac 1 2 m v_1 ^2 + \frac 1 2 m v_2 ^2\\ v^2 &= v_1^2 =v_2^2\\ |\mathbf v|&=|\mathbf v_1| = |\mathbf v_2 | \end{align}

It follows trivially that the initial velocities have reversed their directions and thus the velocities have been exchanged.

The coefficient of restitution is defined as the ratio of relative velocity of separation and relative velocity of approach. Thus

\begin{align} e&=\frac{|\mathbf v_1 -\mathbf v_2|}{|\mathbf v -(-\mathbf v)|}\\ 1&=\frac{|\mathbf v_1 -\mathbf v_2|}{2|\mathbf v|}\\ 2|\mathbf v|&=|\mathbf v_1 -\mathbf v_2| \end{align}

Since this is a one dimensional collision, we can convert the magnitude of difference in velocities to the difference in magnitudes of velocities (with proper sign convention):

$$2v=v_1-v_2$$

Solving this and equation $(1)$ gives us the final velocities. Again you would notice that the initial velocities have reversed and exchanged.

The case which you are thinking about, happens when the collision is perfectly inelastic, which implies that the coefficient of restitution is zero.