The AIC for the piecewise function is indeed the sum of the AIC scores of each piece, so yes, you can just add them together.

Let $X_j$ denote the set of points which "belong" to segment $j$ of your model. The AIC can be derived in the following way. We use $k$ to denote the number of parameters in the model, with $k_j$ being the number of parameters in segment $j$.

\begin{align}
AIC &= 2k-2\log(P(X|\theta))\\
&= 2 \sum_j k_j - 2 \log \prod_j P(X_j | \theta_j) \\
&= \sum_j 2k_j - 2\log P(X_j|\theta_j)\\
&= \sum_j AIC_j
\end{align}

For this proof to go through, there is the caveat that each group $X_j$ must be independent from all other groups in your statistical model. That is, the population growth in each era is independent from all other eras. This is true under a piecewise model of population growth that you are using.