When the $t_i$ are incommensurable in the sense that the $t_i$ generate a dense subgroup, $N(t)\sim CX_0^t$ and this is a consequence of the standard renewal theorem (with no hypothesis on the monotonicity of $N$).

To see this, let $(\xi_n)$ denote some i.i.d. random variables such that $P(\xi_n=t_i)=X_0^{-t_i}$ for every $i$ and 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 $n$, let $S_n=\xi_1+\cdots+\xi_n$. For every $t\ge t_0$, 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 axis, $t_0-(t-S_{T(t)})$ becomes the overshoot over $t-t_0$ for the renewal process based on the sequence $(\xi_n)$ 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 $aX$ is almost surely an integer, hence the non lattice case corresponds to non commensurate parameters $t_i$. 

For  non commensurate parameters $t_i$, all this proves that $M(t)\to E(M(t_0-\xi_0))$, hence
$N(t)\sim CX_0^t$ wih
$$
C=X_0^{-t_0}E(N(t_0-\xi_0)X_0^{\xi_0}).
$$ 
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 the $\xi_n$ given by $P(\xi'=t_i)=t_iX_0^{-t_i}/E(\xi_n)$, hence one can write $C$ as an explicit integral of $N$ over $(0,t_0)$. 

A reference is <a href="http://books.google.com/books?hl=fr&lr=&id=BeYaTxesKy0C&oi=fnd&pg=PR5&dq=Applied+Probability+and+Queues&ots=QxdOx7Pq4a&sig=xYy2TVuwbr-VGz5wb8oxtPP6V3w#v=onepage&q&f=false">Applied Probability and Queues</a> by Søren Asmussen.