How to remove historical values from exponential filter
Yes. Your filter $f(t)$ is an LTI system, so you can use that to your advantage. Let:

$$
C(t) = B(t) - A(t)
$$

Then, if $y_A(t)$, $y_B(t)$, and $y_C(t)$ are the system's responses to $A(t)$, $B(t)$, and $C(t)$ as inputs, then you can straightforwardly do what you want:

$$
A(t) = B(t) - C(t) \rightarrow y_A(t) = y_B(t) - y_C(t)
$$

So, pass $B(t)$ through the filter, then pass $C(t) = [1, 2, 3, 0, 0, 0]$ through the filter and find the difference between the two filter outputs. That's exactly what you would get if you passed $A(t)$ through the filter on its own.