When the $t_i$ are incommensurable in the sense that they generate a dense subgroup, $N(t)=CX_0^t+o(X_0^t)$ for a given constant $C$. This is a consequence of the standard renewal theorem and needs no hypothesis on the monotonicity of the function $t\mapsto N(t)$.

To see this, let $(\xi_k)$ denote some i.i.d. random variables such that $P[\xi_k=t_i]=X_0^{-t_i}$ for every $k$ and $i$. Introduce $M(t)=N(t)/X_0^t$. Then
$$
M(t)=E[M(t-\xi_1)].
$$
Fix $t_0$ such that $t_0\ge t_i$ for every $i$. For every positive $k$, let $S_k=\xi_1+\cdots+\xi_k$. For every $t > t_0$, consider the first time $T(t)$ such that $S_{T(t)}\ge t-t_0$. Since $T(t)$ is a stopping time, the martingale property yields
$$
M(t)=E[M(t-S_{T(t)})].
$$
Reversing the time axis, $t_0-(t-S_{T(t)})$ becomes the overshoot over $t-t_0$ for the renewal process based on the sequence $(\xi_k)$ and starting from $0$. In the non lattice case, the renewal theorem asserts that $t_0-(t-S_{T(t)})$ converges in distribution to a random variable $\xi_0$ when $t\to+\infty$. Being lattice means that there exists a nonzero $a$ such that the random variables $a\xi_k$ are almost surely integer valued, hence the non lattice case corresponds to non commensurate parameters $t_i$.

Thus, when the $t_i$ are non commensurate, $N(t)/X_0^t=M(t)\to C$ wih
$$
C=E[M(t_0-\xi_0)]=X_0^{-t_0}E[N(t_0-\xi_0)X_0^{\xi_0}].
$$
Finally, $\xi_0$ is distributed like $u\xi'$ where $u$ and $\xi'$ are independent, $u$ is uniform on $[0,1]$ and the distribution of $\xi'$ is the size-biased distribution of the distribution of $\xi_k$, given by $P[\xi'=t_i]=t_iX_0^{-t_i}/E[\xi_k]$. Hence one can write $C$ as an explicit integral of the function $N$ over $[0,t_0]$.

A reference is Applied Probability and Queues by Søren Asmussen.