Solution:

The answer is (A). The numbers that can be obtained with a suitable sequence of keys $+1$ and $\times 3$ without two consecutive $+1$'s are all and only those that in base $3$ are written with only digits $0$ and $1$. Indeed, given such a number, to obtain it on the calculator it suffices to scan from left to right the digits of its representation in base $3$ and press $+1$ for the first $1$, then $\times 3$ every time a $0$ is encountered, $\times 3$ followed by $+1$ every time a $1$ is encountered. In base $3$, in fact, multiplying by $3$ means shifting one position to the left. In this way, to each string of $0$'s and $1$'s in base $3$ is associated a sequence of key presses that yields the corresponding number without making the calculator explode. The converse is also true: the sequences of keys involved are all and only those without two consecutive $+1$'s, which would insert a $2$ among the digits of the number. It remains only to count how many numbers with at most $5$ digits in base $10$ are written in base $3$ with only digits $0$ and $1$. We note that $3^{11} > 10^{5}$ while $3^{10} + 3^{9} + \ldots + 3^{0} = \frac{3^{11} - 1}{2} < 10^{5}$, so these are all and only those with at most $11$ digits in base $3$. Since there are $2$ choices ($0$ and $1$) for each digit, such numbers are $2^{11} = 2048$. Note that in this way we have counted both the numbers with $11$ digits (with the first digit equal to $1$), and those with fewer than $11$ digits (which correspond to those with $11$ digits but with a suitable number of leading zeros).