I just want to attempt to clarify something about your question:
Say I posed the question as the oracle lets you know the S(t+1)>S(t)+σ. Then a portfolio would be to buy call options with strike price equal to the current price.
Well actually, no, I wouldn't do this. In this case the precise information possessed is that the stock price at expiry will be greater than the current price by $\sigma$, but you do not know by how much. If you buy a call option with a strike price equal to the current price you have unnecessarily introduced market risk into your profit, i.e. the higher the price goes then the more money you make, but your profit is undetermined.
However, if you go long the market (+1 delta) and sell a call option with a strike at $S(t)+\sigma$ then with certainty you will accrue a profit of the option premium plus $\sigma$. The value of the information is then a fixed deterministic amount. This is a covered call, but equivalently you could also just sell a put option with strike $S(t)+\sigma$, (put-call parity).
Say I posed the question as knowing that |S(t+1)−S(t)|>σ then a portfolio, would be to go and buy a call and a put option with strike price equal to the current price.
Well for the same reason the information gives certainty about a range so, instead of doing a straddle as your comment suggests, I would do a reverse iron butterfly. The value of that information can also be determined as a fixed amount, in that case, with the same profit for all outcomes conditional upon your insider information.
With regards to the question
You will observe that as at time, $t+1$, $|S(t+1)-S(t)|=k$ is a known parameter. Therefore as a trader with flexibility of timing of execution, one particular choice (I am not saying I can prove this is optimal) would be to wait until time $t+1$ and then you have the information about the price at time $S(t+2)$: i.e.
$$a_0 \sigma - k > |S(t+2)−S(t+1)| > a_1 \sigma - k$$
Now you are reduced to some specific cases:
Case 1: $a_1 \sigma - k > 0$ then the reverse iron butterfly can be applied as above (since you have a lower bound of market movement). You also have an upper bound of market movements to further refine your trading window from which you profit by selling a strangle. (executed at time $t+1$ dependent on $k$).
Case 2: $a_1 \sigma - k < 0$ and $a_0 \sigma - k > 0$ then you only have an upper bound, the lower bound is replaced by 0 (so the reverse iron butterfly no longer works) but the upper bound strategy in case 1 will still generate fixed profit by selling a strangle dependent upon $k$.
Case 3: $a_0 \sigma - k < 0$. Well this actually contradicts your piece of information since by definition of modulus $|S(t+2)−S(t+1)| \geq 0 $ hence you can conclude that $k = |S(t+1) - S(t)| \leq a_0 \sigma$, and therefore this opens a trade option at time $t$ with expiry $t+1$ again playing on the fact that you know the price range. So again you sell a strangle here with strikes dependent upon $a_0 \sigma$.
If I was assessing the value of the piece of information I would assess the value of the strangle in case 3 as a fixed amount, and plus the values of cases 1 and 2 integrated over all legitimate values of $k$, which represents the expectation of the informational value from time $t+1$ to $t+2$.
Have only pondered on this for a brief period so I welcome peer review and criticism of these ideas.