The law is of conservation of the line of momentum and its magnitude.
The direction of the line and magnitude are given by linear momentum $\boldsymbol{p} = p \,\hat{\boldsymbol{e}}$.
where $p$ is the scalar magnitude and $\hat{\boldsymbol{e}}$ is the unit direction vector.
The line in space is given by the angular momentum $\boldsymbol{L} = \boldsymbol{r} \times \boldsymbol{p}$.
Where $\boldsymbol{r}$ is the location of any point along the line.
The location of the line is recovered with the following expression $$\boldsymbol{r} = \frac{ \boldsymbol{p} \times \boldsymbol{L} }{ p^2 } \tag{1}$$
So the conservation of angular momentum means conservation of $(\boldsymbol{r} \times \boldsymbol{p})$ and since $\boldsymbol{p}$ is already conserved, it requires that $\boldsymbol{r}$ is conserved also.
So the location of the line is conserved also in addition to the direction and magnitude
Please note that the line of momentum is called the axis of percussion. It signifies the location in space where an impulse needs to be applied in order to completely stop the object. The impact needs magnitude (the impulse) and direction (the contact normal), in addition to a location along its line (the point of contact) to match up with the momentum of the object.
Answers
The total momentum is conserved before and after impact, and by total momentum where I mean a) its magnitude, b) direction, and c) location in space.
Angular momentum of a particle is simply a way to express where the axis of percussion is. And for a particle, the axis of percussion is equivalent to the line of motion.
Addendum
There is a fourth quantity that is conserved and it is called the pitch, defined by $$h = \frac{\boldsymbol{p} \cdot \boldsymbol{L}}{p^2} \tag{2}$$ This is zero for most cases, except for a rigid body that translates and rotates about the same axis. In such a case a single impact cannot completely stop the body.
The angular momentum of a moving rigid body is completely defined by $$ \boldsymbol{L} = \mathrm{I}_{\rm COM} \boldsymbol{\omega} + \boldsymbol{r}_{\rm COM} \times \boldsymbol{p} \tag{3} $$
It can also be completely defined by $$ \boldsymbol{L} = \boldsymbol{r}_{\rm AOP} \times \boldsymbol{p} + h\,\boldsymbol{p} \tag{4}$$
One can go from the first to the second form with expressions (1) and (2) from above to find the location of the percussion axis $\boldsymbol{r}_{\rm AOP}$ and pitch $h$ respectively.
Example
A vertical rod of length $\ell$ and mass $m$ is suspended from one end using a long thin inelastic string from the ceiling. Find where the rod was impacted if after the impact it is exhibiting a pure rotation about the point of suspension.
Set a coordinate frame on the point of suspension and define the following quantities
$$\begin{aligned}
\boldsymbol{r}_{\rm COM} & = \pmatrix{0 \\ -\tfrac{\ell}{2} \\ 0} & \mathrm{I}_{\rm COM} = \left[ \matrix{ \ldots & & \\ & 0 & \\ & & \tfrac{m}{12} \ell^2} \right] \\
\boldsymbol{\omega} & = \pmatrix{0 \\ 0 \\ \dot{\theta}} & \boldsymbol{v}_{\rm COM} = \pmatrix{ \tfrac{\ell}{2} \dot{\theta} \\ 0 \\ 0}
\end{aligned}$$
Then find the linear and angular momentum
$$\begin{aligned}
\boldsymbol{p} & = m \boldsymbol{v}_{\rm COM}
= \pmatrix{m \tfrac{\ell}{2} \dot{\theta} \\ 0 \\} \\
\boldsymbol{L} & = \mathrm{I}_{\rm COM} \boldsymbol{\omega} + \boldsymbol{r}_{\rm COM} \times (m \boldsymbol{v}_{\rm COM}) = \pmatrix{0 \\ 0 \\ \tfrac{m \ell^2}{3} \dot{\theta}} \end{aligned}$$
Finally, find the axis of percussion location
$$ \boldsymbol{r}_{\rm AOP} = \frac{ \boldsymbol{p} \times \boldsymbol{L} }{ \| \boldsymbol{p} \|^2} = \pmatrix{0 \\ -\tfrac{2}{3} \ell \\ 0} $$
So the impact must have happened 2/3 along the rod.
