Denote by $\mu(A_t)$ the max spectral radius of $A_t$ which is defined as the maximal cycle geometric mean
$$ \mu(A_t) := \max \{ (a_{i_1i_2}a_{i_2i_3}\dots a_{i_ki_1})^{1/k}\}$$
where the maximum is taken over all cycles in the matrix $A$, $k$ is the length of the cycle, and for each cycle the indices $i_1,\ldots,i_k$ are distinct (so that it is really a cycle). Here this means that 
$$ \mu(e^t) := \mu(A_{e^t}) = \exp\left( \frac{\ell}{k} t\right)$$
for some maximal cycle of length $k$ and the powers along this cycle sum to $\ell$. In particular, the maximum is attained in the same cycle for all $t\geq0$. 

If $n$ is the size of the matrix, then it is known for any nonnegative matrix $B$ that
$$ \mu(B) \leq \Lambda(B) \leq n \mu(B), $$
(equation (6.10) in Elsner, Johnson, Dias da Silva, The Perron root of a weighted geometric mean of nonnegative matrices. Linear and Multilinear Algebra 24 (1988) 1-13.)

So here this means
$$ \exp\left( \frac{\ell}{k} t\right) \leq \Lambda(e^t) \leq n\exp\left( \frac{\ell}{k} t\right)$$
where the $\ell/k$ is actually easy to get from the matrix $A_t$.