We have
$$S = \sum_{r=0}^n {n+r\choose r} \frac{1}{2^r}
= \sum_{r\ge 0} {n+r\choose n} \frac{1}{2^r}
[[0\le r\le n]]
\\ = \sum_{r\ge 0} {n+r\choose n} \frac{1}{2^r}
[z^n] \frac{z^r}{1-z}
= [z^n] \frac{1}{1-z}
\sum_{r\ge 0} {n+r\choose n} \frac{1}{2^r} z^r
\\ = [z^n] \frac{1}{1-z} \frac{1}{(1-z/2)^{n+1}}
= 2^{n+1} [z^n] \frac{1}{1-z} \frac{1}{(2-z)^{n+1}}
\\ = (-1)^n 2^{n+1} [z^n] \frac{1}{z-1} \frac{1}{(z-2)^{n+1}}.$$
This is
$$(-1)^n 2^{n+1} \mathrm{Res}_{z=0} \frac{1}{z^{n+1}}
\frac{1}{z-1} \frac{1}{(z-2)^{n+1}}.$$
Resides sum to zero. The residue at infinity is zero by inspection.
The residue at $z=1$ is $- 2^{n+1}.$ For for the residue at $z=2$
we require
$$\frac{1}{n!} \left( \frac{1}{z^{n+1}}
\frac{1}{z-1} \right)^{(n)}
\\ = \frac{1}{n!} \sum_{q=0}^n {n\choose q}
(-1)^q \frac{(n+q)!}{n!} \frac{1}{z^{n+q+1}}
(-1)^{n-q} (n-q)! \frac{1}{(z-1)^{n-q+1}}
\\ = (-1)^n \sum_{q=0}^n {n+q\choose q}
\frac{1}{z^{n+q+1}}
\frac{1}{(z-1)^{n-q+1}}.$$
Evaluate at $z=2$ with the factor in front
$$(-1)^n 2^{n+1} (-1)^n
\sum_{q=0}^n {n+q\choose q}
\frac{1}{2^{n+q+1}}
= \sum_{q=0}^n {n+q\choose q}
\frac{1}{2^{q}} = S.$$
This yields
$$S - 2^{n+1} + S = 0$$
or
$$\bbox[5px,border:2px solid #00A000]{
S = 2^n.}$$
This problem is a special case of MSE
538309.
Addendum.
An alternate approach uses
$$S = \sum_{q=0}^n {n+q\choose n} \frac{1}{2^q}
= \sum_{q=0}^n \frac{1}{2^q} [z^n] (1+z)^{n+q}
= [z^n] (1+z)^n \sum_{q=0}^n \frac{1}{2^q} (1+z)^q
\\ = [z^n] (1+z)^n
\frac{1-(1+z)^{n+1}/2^{n+1}}{1-(1+z)/2}
= [z^n] (1+z)^n
\frac{2-(1+z)^{n+1}/2^{n}}{1-z}
\\ = 2\times 2^n
- [z^n] (1+z)^{2n+1} \frac{1}{2^n} \frac{1}{1-z}
\\ = 2\times 2^n
- \frac{1}{2^n} \sum_{q=0}^n {2n+1\choose q}
= 2\times 2^n - \frac{1}{2^n} \frac{1}{2} 2^{2n+1}
= 2^n.$$
