The simplest way to compute the requested area would be the following:
The axis of the cylinder $S$ is parallel to the $x$-axis, and $S$ intersects the $(y,z)$-plane in the circle $y^2+(z-1)^2=1$ of unit radius. A parametric representation of this circle is given by
$$y=\sin\phi,\quad z=1+\cos\phi\qquad(-\pi\leq \phi\leq\pi)\ .$$
Through each point on this circle passes a stalk $s_\phi$ parallel to the $x$-axis whose ends are determined by
$$x^2=y^2+z^2=2(1+\cos\phi)\ .$$
It follows that $s_\phi$ has length
$$\ell(\phi)=2\sqrt{2(1+\cos\phi)}=4\cos{\phi\over2}\qquad(-\pi\leq \phi\leq\pi)\ .$$
The area $\omega$ of the surface in question is therefore given by
$$\omega=\int_{-\pi}^\pi\ell(\phi)\ d\phi=8\int_{-\pi}^\pi\cos{\phi\over2}\ d\phi=16\ .$$
First we parameterize our surface by two families of curves $\{u=c\}$ and $\{v=d\}$:
The we can see that $$\mathbf r_u = \frac{\partial \mathbf r(u_0,v_0)}{\partial u} = \lim_{h\to 0} \frac{\mathbf r(u_0+h,v_0)-\mathbf r(u_0,v_0)}{h}$$
will be tangent to the curve $u=u_0$ at the point $(u_0,v_0)$ and thus also tangent to the surface. Likewise for $\mathbf r_v = \dfrac{\partial \mathbf r(u_0,v_0)}{\partial v}$.
Also notice that because of the way we choose our charts $\mathbf r_u$ and $\mathbf r_v$ will point in different directions. Thus they will be a basis for the tangent space at that point.
Then we realize that, in analogy to how every differentiable curve looks locally like a straight line, every surface looks locally like a plane:
And we know how to find the area of a planar region: we chop it into looks rectangles (or parallelograms) and find the area of each of those.
So that's what we do in the tangent space.
In the above image, we see a $\mathbf r(u_0 + \delta u, v_0) - \mathbf r(u_0,v_0)$. Using what we know of single variable calculus, partial derivatives, and vectors we can see that this is just $\mathbf r_u(u_0,v_0)du$. Likewise the other side of that parallelogram is $\mathbf r_v(u_0,v_0)dv$. Then to find the area of that we just take the norm of the cross product:
$$dA = \|\mathbf r_u(u_0,v_0)du\times \mathbf r_v(u_0,v_0)dv\| = \|\mathbf r_u(u_0,v_0)\times \mathbf r_v(u_0,v_0)(dudv)\| \\ = \|\mathbf r_u(u_0,v_0)\times \mathbf r_v(u_0,v_0)\||dudv| = \|\mathbf r_u(u_0,v_0)\times \mathbf r_v(u_0,v_0)\|dudv$$
where I was able to move the $du$ and $dv$ around and outside the norm because they are (positive) scalar differentials.
Then finally, we know that the surface area will just be the integral of all of the differential area elements:
$$A = \int_D dA = \int_D \|\mathbf r_u\times \mathbf r_v\|dudv$$
and we have our expresssion for surface area of a parameterized surface.
Best Answer
There are two cylinders $A$ and $B$. The surface of $A\cap B$ consists of two parts:
You dealt with only one of the above. So, after doubling your half-cylinder to cylinder, you should double again.