Let $M$ be the middle of $SC$. Then $|SM| = |MC| = 1$
Because $|AC| = |AS|$, we have that $AMS$ is a right triangle, so by the Pythagorean theorem we have $|AM| = \sqrt{|AS|^2 - |MS|^2} = \sqrt{2^2 - 1} = \sqrt3$, similarly $|BM| = \sqrt3$.
But then $|AM|^2 + |BM|^2 = 3 + 3 = 6 = |AB|^2$, so by the converse of the Pythagorean theorem the triangle $AMB$ is right.
So the volume can be calculated as (base $ACS$ and height $MB$) $$V = \frac13\cdot\frac12|CS||MA|\cdot|MB| = \frac16\cdot2\cdot\sqrt3\cdot\sqrt3 = \frac13\cdot3 = 1$$