I believe your question is still open. Some notation before we move on to a related conjecture:
Denote $a\preceq b$ if $a$ is a subsequence of $b$ when they are both written in base $10$, e.g., $553\preceq 65536$, $63\preceq 65536$, but $35$ is not a subsequence (so order matters). Given a set $S\subseteq\mathbb{N}$, we define the minimal set as the smallest set $M\subset S$, such that for all $s\in S$, there exists $m\in M$ and $m\preceq s$. From Higman's lemma, we know that the minimal set is finite for every possible $S$.
Back to your question, there is a weaker conjecture by J. Shallit in
Minimal Primes, J. Recreational Math. 30 (2000), 113-117
that $\{1,2,4,8,65536\}$ is the minmal set of powers of $2$ in base $10$, which as far as I can tell is still open. As Shallit observed in the paper that this conjecture is true if "every power of $16$ greater than $65536$ contains a digit in the set $\{1,2,4,8\}$", which is precisely your question.
For numerical search, you can take a look at $A137998$. The question is equivalent to the statement that $16^4$ is the only entry of $0$ that appears in this sequence. Due to a repeating pattern mentioned in its comments, you would only need to check at most $12$ out of every $25$ powers of $16$.
(Note this had been posted as a comment, but as it was the 10+th comment, I know that not everyone would notice it. I hope it has been expanded enough to be warranted as an answer. Please let me know if otherwise)
edited: below is taken from Shallit's paper as an attempt to make the concept of "minimal set" more accessible.
We will take a look at prime numbers. If I were to claim, as an analogy to the original question, that all prime numbers when written in base-$10$ must contain a prime digit, i.e., $\{2,3,5,7\}$, you would easily be able to find a counterexample: "How about $11$? and $19$?". However, I could expand the list to include all $2$-digit prime numbers that has no prime digit, i.e., $\{2,3,5,7,11,19,41,61,89\}$. Then I would claim that all prime numbers must contain a prime number in this new list. This is where I would need to clarify what it means by a number to "contain" another number with more than one digit, e.g., $19$. The defintion chosen here is that given a number $b$, if we can reach $a$ by deleting some digits of $b$, then we say $b$ contains $a$ (or $a\preceq b$). As an example, $109$ is a prime number and it contains $19$. This time it would take you a bit effort to find a disproof: the first one is $409$. The procedure goes on as I would extend the list to include such $3$-digit prime numbers and you would produce larger counter-examples, so on so forth.
The surprise is that this process necessarily ends, due to Higman's lemma. As my list grows longer (but remains finite), at certain point the claim becomes true. We call the final list minimal set of prime numbers.
We can repeat this process for powers of $2$, which leads to the conjecture that $\{1,2,4,8,65536\}$ is the minimal set. This is apparently weaker than your question, as it allows the possibility that there exists some power of $2$ that does not contain $1,2,4,8$ but contains $65536$. Note the proofs we know for Higman's lemma, albeit constructive, are not effective to give us the actual minimal set in general. Shallit did manage to get the answer for prime numbers using elementary means, but as far as I know, there is no progress for the minimal set of powers of $2$, which in turn indicates that the stronger statement in your question is still open.
Best Answer
The idea, as Greg Martin says in the comments, is that you can ignore most of the digits and compute using only the first $10 + k$ digits for some $k$. Since we consider a sum of $100$ numbers, the sum of the errors if we discard all of the digits after the first $10 + k$ is less than $100 \times 10^{40-k}$, so if we pick $k = 2$ the sum of the errors is less than $10^{40}$ and so does not affect the first $10$ digits. So, as Greg says, it suffices to keep the first $12$ digits of each number. This is true regardless of how long they are, the important thing is that we are only summing $100$ numbers. If we were summing $1000$ numbers we'd need to keep the first $13$ digits, etc.