Inspired strongly by Anthony's answer, here is a formula that works for arbitrary $A$. Let $M$ be the $n^2 \times n^2$ square matrix satisfying $$M_{ij,kl}= A_{i,k}+A_{j,l}$$ Then (because the two summands commute) $$(e^{M t})_{ij,kl} = (e^{A t})_{i,j} (e^{At})_{k,l}$$ so in particular $$(e^{At})_{i,j}^2 = (e^{Mt})_{ii,jj}$$ and $$\int_t (e^{At})_{i,j}^2 = \int_t (e^{Mt})_{ii,jj} = - (M^{-1})_{ii,jj}$$

(Here I am using commas to separate the two indices of a matrix entry)

Inspired strongly by Anthony's answer, here is a formula that works for arbitrary $A$. Let $M$ be the $n^2 \times n^2$ square matrix given by $$M= A \otimes I_n + I_n \otimes A_n$$ i.e. in terms of indices

$$ M_{ij,kl}=A_{i,k} \delta_{j,l} + \delta_{i,k}A_{j,l}$$

Then because $A \otimes I_n$ and $I_n \otimes A$ commute, $$e^{Mt} = (e^{At} \otimes I_n) (I_n \otimes e^{At}) = e^{At} \otimes e^{At}$$ i.e. in indices $$(e^{M t})_{ij,kl} = (e^{A t})_{i,j} (e^{At})_{k,l}$$ so in particular $$(e^{At})_{i,j}^2 = (e^{Mt})_{ii,jj}$$ and $$\int_t (e^{At})_{i,j}^2 = \int_t (e^{Mt})_{ii,jj} = - (M^{-1})_{ii,jj}$$

(Here I am using commas to separate the two indices of a matrix entry)