This is known as Archimedes' Hat-Box Theorem.
Archimedes developed the methods for solving such problems
many centuries before the invention of calculus,
so I suppose they would suffice for a "non-calculus" answer.
There is a discussion of this theorem (along with some nice
three-dimensional diagrams) on Zachary Abel's Math Blog, under the heading "Spherical Surfaces and Hat Boxes."
The method described on that page uses the fact that if we have some
region $A$ on the surface of a sphere, the volume of the
interior part of the sphere between the sphere's center and $A$ is
$\frac13$ the radius of the sphere multiplied by the area of $A$.
This can be shown through the method of exhaustion by triangulating
$A$ with finer and finer meshes and considering the volumes of
triangular pyramids (determined by the meshes)
whose bases are inside or outside the sphere.
I think it is a bit simpler to prove the theorem if we assume
initially that one of the planes is tangent to the sphere;
that is, the part of $\mathcal S$ that we measure is a spherical cap.
(To get the more general result,
if $-1 < a < b < 1$ subtract the spherical cap
above the plane $x_3 = b$ from the spherical cap above $x_3 = a$.)
For a plane $x_3=a$ cutting a unit sphere $\mathcal S$,
we construct two three-dimensional regions.
The first region, $U$, is bounded by the spherical cap cut off by the plane
and the cone whose vertex is the center of the sphere and whose base
is the circular boundary of the spherical cap.
The second region, $V$, is the region bounded by the cylinder
by the cone $x_1^2+x_2^2=x_3^2$,
and by the cone $a^2 x_1^2 + a^2 x_2^2 = x_3^2$.
That is, $V$ consists of all points between the center of the sphere and
the part of the cylinder $x_1^2+x_2^2=1$ between $x_3 = a$ and $x_3 = 1$.
Now consider a horizontal plane $x_3 = z$, where $0 < z < 1$.
The intersection of this plane with $U$ is a circle, while the intersection
of the same plane with $V$ is an annulus (a disk with a circular hole).
Moreover, a little geometry shows that for any given $z$,
the area of the circle is always equal to the area of the annulus.
Since this applies to every simultaneous cross-section of $U$ and $V$
by any plane perpendicular to the $x_3$ axis,
by Cavalieri's principle, the volumes of $U$ and $V$ are equal.
We can show that the volume of $V$ is $\frac13$ of the area of the
part of the cylinder $x_1^2+x_2^2=1$ between $x_3 = a$ and $x_3 = 1$
(one way is to add the volume inside the cylinder
between the planes $x_3 = a$ and $x_3 = 1$
to the volume of the cone with vertex at $0$ and base
$x_1^2+x_2^2\leq 1, x_3 = a$
and subtract the volume of the cone with vertex at $0$ and base
$x_1^2+x_2^2\leq 1, x_3 = 1$).
But the volume of $U$ is $\frac13$ the area of the spherical cap
(since the radius of the sphere is $1$). Since this is equal to the
volume of $V$, the area of the spherical cap must equal the area of
the cylinder $x_1^2+x_2^2=1$ between $x_3 = a$ and $x_3 = 1$.
An alternative method is to take the zone of the sphere between
the two planes $x_3=z$ and $x_3=z+\delta$,
approximate the area of this zone by the area of a frustum
(truncated cone) between the circles
$x_1^2+x_2^2=x_3^2, x_3z$ and $x_1^2+x_2^2=x_3^2, x_3=z+\delta$,
and show that this area is the area of the cylinder
$x_1^2+x_2^2=1$ between the planes $x_3=z$ and $x_3=z+\delta$.
I think a little trigonometry is generally used to approximate
the area of the frustum; the approximation over a finite number of
frustums converges to the exact area of the spherical cap as we
reduce the spacing between the parallel planes.
A possible third method would be to inscribe a spherical polygon inside
a spherical cap and circumscribe another spherical polygon around the
spherical cap. The areas of the polygons are the sums of areas of
spherical triangles. The area of the cap would be the area that the
circumscribed and inscribed polygons converge to, by principle of exhaustion.
If you can find the area of intersection of a spherical cap and a lune
with vertex at the center of the spherical cap, you can get the area
of the cap directly, but while the intersection of the cap and lune
is "triangular" in some sense, I would not consider it
a true spherical triangle, since one of the "edges" does not lie
along a great circle.