Yes. If we know any four consecutive terms of $S$, we can calculate "the whole sequence", i.e. we know the terms after these four, as well as those before. Since there are only finitely many combinations of four digits, that means that some combination will repeat itself. Obviously, somewhere between those two occurences, digits $1, 9, 8, 2$ come up again. By calculating the four terms before $1, 9, 8, 2$, we see that those are exactly $3, 0, 4, 4$ ($x_4 + 1 + 9 + 8\equiv 2 \,(\cdot10)$ implies $x_4 = 4$, $x_3 + 4 + 1 + 9\equiv 8 \, (\cdot 10)$ implies $x_3 = 4$, $x_2 + 4 + 4 + 1\equiv 9 \, (\cdot 10)$ yields $x_2 = 0$, and finally $x_1 + 0 + 4 + 4\equiv 1 \, (\cdot 10)$ yields $x_1 = 3$). We conclude that digits $3, 0, 4, 4, 1, 9, 8, 2$ come up consecutively in $S$.

Could you please explain a little more this? idioteque wrote: that means that some combination will repeat itself. Obviously, somewhere between those two occurences, digits $1, 9, 8, 2$ come up again.

Going backwards from the first occurence of that 4-digit combination, we will get to the beginning of this sequence, i.e. to $1, 9, 8, 2$. That means that going backwards from the second occurence of that 4-digit combination we will also get to $1, 9, 8, 2$, since we use the same rule to find the digits before... This actually means this sequence is periodic. Let's say that digits $0, 8, 6, 3$ (in that order) repeat themselves. Going back from the first occurence of it, we find that the members before are (I write them in the order we calculate them) $9, 9, 0, 2, 8, 9, 1$. We see that we came to $1, 9, 8, 2$, which means that $1, 9, 8, 2$ will also appear if we calculate terms before second occurence of $0, 8, 6, 3$. Regardless of that "repeating" combination, going back from the first occurence of it, we will get to $1, 9, 8, 2$ (beginning of the sequence), so we know that $1, 9, 8, 2$ appear somewhere between the first two occurences of that repeating combination. I hope it's somewhat clearer...