Let $X_1,X_2,\dots$ be independent identically distributed random variables representing the successive jumps of the particle. Let $M_n$ be the position of the particle at time $n\in\{0,1,\dots\}$, so that $M_0=0$ and $M_n=\max(0,M_{n-1}+X_n)$ for $n\in\{1,2,\dots\}$. Let now $F_n(x):=\mathbb{P}(M_n\le x)$ for real $x$. Then $F_0(x)=\mathrm{I}\{x\ge0\}$ and $F_n(x)=\frac{\mathrm{I}\{x\ge0\}}{1+r}\,\int_{-r}^1 F_{n-1}(x-y)\,dy$ for all real $x$ and $n\in\{1,2,\dots\}$. In particular, the probability in question is $P_{n,r}=\mathbb{P}(M_n=0)=F_n(0)$. 

The work in Mathematica notebook at [n=1,2,3][1] (the corresponding pdf file at [n=1,2,3 - pdf][2]) shows that the expression 
$$\frac{\left(\frac{3r}{1+r}\right)_{n-1}}{(1+r)\left(2r\right)_{n-1}},\quad r\geq 1,$$
which you most recently proposed for $P_{n,r}$, is incorrect already for $n=3$ (and $r\ne1$), even though it seems pretty close to the correct one.  

I doubt that there is such a simple closed form expression for $P_{n,r}$. However, using Spitzer's identity and work done in [c.f. of X_+][3], one can obtain an integral expression for the generating function of $P_{n,r}$: 
$$
(*)\qquad \sum_{n=0}^\infty P_{n,r}z^n=
\frac1{\sqrt{1-z}}\exp \Big(\frac1{2 \pi  i}\,\int_0^{\infty } 
\ln\frac{(r+1) u-z (e^{i r u}-e^{-i u})}{(r+1) u-z (e^{i u}-e^{-i r u})} \, \frac{du}{u}\Big)
$$ 
for complex $z$ with $|z|<1$. 
To obtain (*), one can start with formula (23) in [c.f. of X_+][4], analytically extend it to complex $s$ with $\tau:=\Im s\ge0$, and then let $\tau\to\infty$. Alternatively, one can start with the second displayed formula on page 159  (with $\lambda=0$) in 
[Spitzer 1960][5] and then reason as in [c.f. of X_+][6]; here one should also have in mind statement (2.4) on page 155 in [Spitzer 1960][7]. 

In particular, it easily follows from (*) that your expression for $P_{n,r}$ is correct in the symmetric case, when $r=1$, but apparently only in this case. 

**Addendum:** Even though there does not seem to be a simple closed form expression for $P_{n,r}$, an explicit but rather complicated expression for $P_{n,r}$ can be obtained from (*). Indeed, expand the logarithm back in powers of $z$. Then use the Cauchy integral formula to see that 
$$\frac1{\pi i}\int_0^\infty\frac{f(u)^n-f(-u)^n}u\,du =1-2a_n,$$
where 
$$f(u):=\mathbb{E}e^{iuX_1}=\frac{e^{iru}-e^{-iu}}{i(r+1)u}$$
for $u\ne0$ and 
$$a_n:=a_{n,r}:=\frac1{2n}\Big[1+\frac1{n!}\sum_{j=0}^n(-1)^j \binom nj \Big(\frac n{r+1}-j\Big)^n\,\text{sign}\Big(\frac n{r+1}-j\Big)\Big].$$ 

It follows from (*) that for complex $z$ with $|z|<1$ 
$$\sum_{n=0}^\infty P_{n,r}z^n=\exp\sum_{k=1}^\infty a_k z^k
=\prod_{k=1}^\infty\exp(a_k z^k)
=\prod_{k=1}^\infty\sum_{q=0}^\infty \frac{a_k^q z^{kq}}{q!},$$
whence 
$$P_{n,r}=\sum\prod_{k=1}^n\frac{a_k^{q_k}}{q_k!},$$
where the sum is taken over all $n$-tuples $(q_1,\dots,q_n)$ of nonnegative integers such that $1q_1+2q_2+\dots+nq_n=n$. In particular, for $n=0$ the set of all such $n$-tuples is the singleton set $\{\emptyset\}$, and, as usual, $\prod_{k=1}^0\ldots:=1$, so that $P_{0,r}=1$. Also, 
$$P_{1,r}=a_1=a_{1,r},\quad P_{2,r}=a_2+a_1^2/2!, \quad P_{3,r}=a_3+a_1a_2+a_1^3/3!. $$
Substituting here the expressions for the $a_k$'s, one sees that the above results for $n=1,2,3$ agree with the ones previously found in the Mathematica notebook at [n=1,2,3][1] (the corresponding pdf file at [n=1,2,3 - pdf][2]) by iterative integration. 


  [1]: https://www.dropbox.com/s/w6w2obndgxcdr0y/walk%3Bn%3D1%2C2%2C3.nb?dl=0
  [2]: https://www.dropbox.com/s/3c3j39hntqtaitf/walk%3Bn%3D1%2C2%2C3.pdf?dl=0
  [3]: http://arxiv.org/abs/1309.5928v3
  [4]: http://arxiv.org/abs/1309.5928v3
  [5]: http://www.ams.org/journals/tran/1960-094-01/S0002-9947-1960-0111066-X/
  [6]: http://arxiv.org/abs/1309.5928v3
  [7]: http://www.ams.org/journals/tran/1960-094-01/S0002-9947-1960-0111066-X/