We start off by observing that in order to get from $x$ to $0$, we first have to pass through $x-1$; the time to get from $x$ to $0$ will be equal to the first-passage time from $x$ to $x-1$ + the time from $x-1$ to 0, and obviously this applies to expectations as well.
Applying this logic recursively establishes the following relationship:
$$\mathbb{E}(\tau|S_0=x) = x\mathbb{E}(\tau|S_0=1)$$
Now for $\mathbb{E}(\tau|S=1)$. If we are at $S=1$, we have a probability $1-p$ of reaching $0$ in one step and a probability $p$ of transitioning to $S=2$, at which time (now time $1$ instead of time $0$) we have an expected time-to-$0$ of $2\mathbb{E}(\tau|S=1)$:
$$\mathbb{E}(\tau|S=1) = (1-p) + p(1 + 2\mathbb{E}(\tau|S=1))$$$$\mathbb{E}(\tau|S=1) = (1-p)\cdot 1 + p(1 + 2\mathbb{E}(\tau|S=1))$$
Rearranging terms gives us:
$$\mathbb{E}(\tau|S=1) (1-2p) = (1-p) + p = 1 $$
$$\mathbb{E}(\tau|S=1) = {1 \over 1-2p}$$
Combining this with our first equation gives us the desired result:
$$\mathbb{E}(\tau|S_0=x) = {x \over 1-2p}$$
The variance can be found in a similar way.