No, there's no irrational $s$ with this property. Here's a concrete hands-on argument. (This may be equivalent to Ville Salo's comment, but I'm not familiar with that terminology.)

The sequence $d = s\oplus \ell_1(s)$ is just the sequence of differences (mod 2) between successive elements of $s$. Given $s(0)$ we can reconstruct all of $s$ from $d$, since $$s(k) = s(0) \oplus \bigoplus_{i=0}^{k-1} d(i)$$ If $d$ has period $n$, then $s$ must be periodic, since $$s(2n) = s(0) \oplus \bigoplus_{i=0}^{2n-1} d(i) = s(0) \oplus 2 \bigoplus_{i=0}^{n-1} d(i) = s(0)$$

No, there's no irrational $s$ with this property. Here's a concrete hands-on argument. (This may be equivalent to Ville Salo's comment, but I'm not familiar with that terminology.)

The sequence $d = s\oplus \ell_1(s)$ is just the sequence of differences (mod 2) between successive elements of $s$. Given one entry in the sequence, $s(j)$, we can reconstruct the rest of $s$ from $d$, since: $$s(j + 1) = s(j) \oplus d(j)$$ So if $d$ has period $n$, then $s$ must have period no more than $2n$, since $$s(j + 2n) = s(j) \oplus \bigoplus_{i=0}^{2n-1} d(i) = s(j) \oplus 2 \bigoplus_{i=0}^{n-1} d(i) = s(j)$$

No, there's no irrational $s$ with this property. Here's a concrete hands-on argument. (This may be equivalent to Ville Salo's comment, but I'm not familiar with that terminology.)

The sequence $d = s\oplus \ell_1(s)$ is just the sequence of differences (mod 2) between successive elements of $s$. Given $s(0)$ we can reconstruct all of $s$ from $d$, since $$s(k) = s(0) \oplus \bigoplus_{i=0}^{k-1} d(i)$$ If $d$ has period $n$, then $s$ must be periodic, since $$s(2n) = s(0) \oplus \bigoplus_{i=0}^{2n-1} d(i) = s(0) \oplus 2 \bigoplus_{i=0}^{n-1} = s(0)$$

No, there's no irrational $s$ with this property. Here's a concrete hands-on argument. (This may be equivalent to Ville Salo's comment, but I'm not familiar with that terminology.)

The sequence $d = s\oplus \ell_1(s)$ is just the sequence of differences (mod 2) between successive elements of $s$. Given $s(0)$ we can reconstruct all of $s$ from $d$, since $$s(k) = s(0) \oplus \bigoplus_{i=0}^{k-1} d(i)$$ If $d$ has period $n$, then $s$ must be periodic, since $$s(2n) = s(0) \oplus \bigoplus_{i=0}^{2n-1} d(i) = s(0) \oplus 2 \bigoplus_{i=0}^{n-1} d(i) = s(0)$$