If the sequence contains a power of $2$, easy to see that $5$ does not divide $a_1$. Suppose assume the latter. Then, note that for any natural $t$, we have some $k$ with $\left\lfloor{\frac{a_k}{10\right}\rfloor=t}$. Suppose $2^n=10x+8$ does not occur in the sequence, it means there exists $u$ with $a_u=10x+6$ but then easy to see that $2^{n+2}=a_{u+6x+5}$ and so, we have infinitely many powers of $2$ in the sequence.

The ''only if'' part is trivial. If $5 \nmid a_1$, its easy to see that we end up at some $t$ such that unit digit of $a_t$ is $4$. Then $a_{t+4} = a_t + 4+ 8 + 6 + 2 = a_t + 20$. Consider the sub sequence $a_t , a_{t+4} , a_{t+8} , \ldots $. This is an arithmetic progression with difference $20$. If $a_t \equiv 4 \mod 20$, that progression contains infinitely many power of two. If $a_t \equiv 14 \mod 20$ , then consider the arithmetic progression (with difference 20) $a_{t+1} , a_{t+5} , \ldots$. Here $a_{t+1} \equiv 8 (\mod 20)$, this contains infinitely many power of two.

This is a old problem.(Maybe 1993 Russian) And has solved here for many times, like http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=443310&hilit=infinitely+many+powers+of+2

If $5|a_1$ then the sequence contains only 2 members, so it can't contain infinitely many powers of 2. Assume now that $5\nmid a_1$. In this case one of the terms of the sequence is congruent to 2 modulo $10$, since $a_{n+1}=a_n+b_n\equiv 2a_n(\text{mod}\ 10)$ for all positive integers $n$. Assume $a_k\equiv 2(\text{mod}\ 10)$. It folows that $a_{k+4s}=a_k+(2+4+8+6)s=a_k+20s$ for all positive integers $s$. Now observe that any power of $2$ can leave one of the remainders from the set $\{2,4,8,16,12\}$ when divided by 20 and both 2 and 12 are in this set. Hence the sequence contains infinitely many powers of $2$.