Consider a continuous time biased random walk on a 1D lattice. The random walker walks with rate $k_\mathrm{R}$ to the right and with rate $k_\mathrm{L}$ to the left. The probability $p_n(t)$ of being at location $n$ at time $t$ is then described by $$\frac{\mathrm{d}}{\mathrm{d}t}p_n(t)=k_\mathrm{R}p_{n-1}(t)+k_\mathrm{L}p_{n+1}(t)-(k_\mathrm{L}+k_\mathrm{R})p_n(t).$$ The solution of this equation is $$p_n(t)=\left(\frac{k_\mathrm{R}}{k_\mathrm{L}}\right)^{\frac{n}{2}}I_n(2\sqrt{k_\mathrm{L}k_\mathrm{R}}t)\mathrm{e}^{-(k_\mathrm{L}+k_\mathrm{R})t},$$ where $I_n(t)$ is the Bessel function.
On the other hand, since this process is related to drift diffusion, we expect the long time limit $t\to\infty$ to be given by $$p_n(t)\sim\frac{1}{\sqrt{4\pi Dt}}\mathrm{e}^{-\frac{(n-vt)^2}{4Dt}},$$ with the "diffusion constant" $D=(k_\mathrm{L}+k_\mathrm{R})/2$ and "drift velocity" $v=k_\mathrm{R}-k_\mathrm{L}$. Visually, this also seems to be correct for all values I tried. Is it possible to derive the asymptotic limit from our expression of $p_n(t)=\left(\frac{k_\mathrm{R}}{k_\mathrm{L}}\right)^{\frac{n}{2}}I_n(2\sqrt{k_\mathrm{L}k_\mathrm{R}}t)\mathrm{e}^{-(k_\mathrm{L}+k_\mathrm{R})t}$?
I have already tried using the integral representation of the Bessel function and applying the saddle-point approximation to it, but this does not give the desired result.
I think that you skipped some steps when doing you did the approximations in the unbiased case. If you do it rigorously, the same method applies. More formally, you want to estimate large deviations at large times, so you want to scale: $$ n = ut $$ You then get: $$ \begin{align} p_n(t) &= \int_0^{2\pi}e^{ikn}\exp\left[-t\left(k_L(1-e^{ik})+k_R(1-e^{-ik})\right)\right]\frac{dk}{2\pi}\\ &= \int_0^{2\pi}\exp\left[t\left(iku-k_L(1-e^{ik})-k_R(1-e^{-ik})\right)\right]\frac{dk}{2\pi}\\ \end{align} $$ The saddle point is the solution $k$ to: $$ u=k_Re^{-ik}-k_Le^{ik} $$ The solution is purely imaginary, and it is better to rewrite it as $k=-i\kappa$. Explicitly: $$ \kappa=\ln\left(\frac{\sqrt{u^2+4k_Rk_L}-u}{2k_L}\right) $$ but it is easier to think in terms of $\kappa$. Therefore, the logarithmic leading order is: $$ \begin{align} \frac1t\ln p_n(t)&\to iku+k_Re^{-ik}+k_Le^{ik}-k_R-k_L\\ &= \kappa(k_Re^{-\kappa}-k_Le^\kappa)+k_Re^{-\kappa}+k_Le^\kappa-k_R-k_L \end{align} $$ The rate function reaches its maximum at: $$ \kappa=0 $$ that is: $$ u=k_R-k_L:=v $$ Taylor expanding for $\kappa\to0$: $$ \lim_{t\to\infty}\frac1t\ln p_n(t)\sim -\frac{k_R+k_L}{2}\kappa^2 $$ Substituting with $u$ and Taylor expanding the exponent about the saddle point for the gaussian integral to get the asymptotic equivalent of $p_n$, you recover your formula.
Hope this helps.