I believe that leonbloy's comment/hint is relevant, and whenever $\alpha$ is rational, $P(\alpha)$ is algebraic. For instance, $P(1/3)$ is simply the probability that a random walk on $\mathbb{Z}$ starting at the origin and taking steps of $+2$ or $-1$ with equal probability will ever reach $-1$. If $f(n)$ is the probability of ever reaching a negative point given that the walk is currently at $n$, then $f(n)$ satisfies $$f(n) = \frac{f(n+2)+f(n-1)}2.$$ The standard ansatz $f(n) = x^n$ gives three solutions for $x$: $x=1$ or $x=(-1\pm \sqrt{5})/2$. It is easily seen that the only solution of the form $f(n) = Ax_1^n + Bx_2^n + Cx_3^n$ that satisfies the boundary conditions (at $-1$ and infinity) is $f(n) = (-1/2+\sqrt{5}/2)^{n+1}$, from which it follows that $$P(1/3) = \frac{\sqrt{5}-1}2.$$ Similarly, $P(1/2)$ is equal to the unique root of $x = (x^4+1)/2$ in $(0,1)$, and in general, $P(k/(k+2))$ is the unique relevant root of $x = (x^{k+2}+1)/2$.
If $\alpha$ is rational but not of this form, the boundary conditions become a little more complicated. For instance, consider $\alpha=1/5$. This can be modeled by a random walk on $\mathbb{Z}$ where a particle takes steps of $+3$ or $-2$. The corresponding ansatz gives $x^2 = (x^5+1)/2$, which has two roots of absolute value smaller than 1, one positive and one negative. Since the walk can now jump to the left, it can reach a first negative value both at $-1$ and at $-2$. Apart from $f(n)\to 0$ at infinity, we get the two boundary conditions $f(-1)=1$ and $f(-2)=1$. We can now find $f$ explicitly (at least numerically) as $f(n) = Ax_1^n+Bx_2^n$ where $x_1$ and $x_2$ are the roots in $(-1,1)$ and $A$ and $B$ are determined by the boundary conditions.
As has already been pointed out, $P(\alpha)$ makes a jump at every rational number. With the approach outlined above, one can in principle compute both the "lower" and "upper" values of $P(\alpha)$ (the upper value being the probability that $S_n$ reaches, but does not cross, the line of slope $\alpha$) whenever $\alpha$ is rational. This is feasible only when $\alpha$ is a relatively simple fraction, but it should still be possible to obtain a good plot of $P(\alpha)$ as a function of $\alpha$.
I recall that after this problem was discussed at the open problem session of FPSAC 2003, Pontus von Brömssen made some such plots. I haven't been in touch with him in the last few years, but apparently he has an (inactive) MO-account. I will notify him of this question.