$\newcommand\Ga\Gamma\newcommand\Z{\mathbb Z}$Let $t:=\lambda$ and $k:=x\in\Z\cap[0,t)$. Then for $X_t\sim Pois(t)$ we have 
$$P(X_t\ge k)=1-\frac{\Ga(k,t)}{\Ga(k)};$$
this can be shown e.g. by integrating 
$$\Ga(k,t)=\int_t^\infty u^{k-1}e^{-u}\,du$$
$k-1$ times by parts. 

It follows now by Corollary 3 of [this paper][1] or its [arXiv preprint version][2] that $P(X_t\ge k)>1/2$. Therefore (as was noted by user J.), standard large-deviation techniques are not applicable here. 

However, using Theorem 1.1 and Proposition 2.7 of [this paper][3] or its [arXiv preprint version][4], we have 
$$P(X_t\ge k)\le B_1(k,t):=1-e^{-t} \Big(1+\frac{(t+2)^k-t^k-2^k}{2 k!}\Big)$$
for $k\ge3$ and 
$$P(X_t\ge k)\le B_2(k,t):=1-\frac{e^{-t} t^{k-1}}{k!}$$
for $k\ge1$. 

Both upper bounds, $B_1(k,t)$ and $B_2(k,t)$, on $P(X_t\ge k)$ are nontrivial, in the sense that they both are strictly less than $1$. Both bounds are rather accurate. 
The bound $B_2(k,t)$ on $P(X_t\ge k)$ is simpler but less accurate than $B_1(k,t)$. What has been said in this paragraph is illustrated by the following graphs 
$\Big\{\Big(k,\dfrac{P(X_{2k}\ge k)}{B_1(k,2k)}\Big)\colon k\in\{3,\dots,20\}\Big\}$ (red) and $\Big\{\Big(k,\dfrac{P(X_{2k}\ge k)}{B_2(k,2k)}\Big)\colon k\in\{3,\dots,20\}\Big\}$ (blue):  

[![enter image description here][5]][5]

Here the maximum relative error of the bound $B_1(k,2k)$ on $P(X_{2k}\ge k)$ is $\approx0.001$ and the maximum relative error of the bound $B_2(k,2k)$ on $P(X_{2k}\ge k)$ is $\approx0.050$. 


  [1]: https://www.sciencedirect.com/science/article/abs/pii/S0167715220303308
  [2]: http://arxiv.org/abs/2012.13578
  [3]: http://mia.ele-math.com/23-95/Exact-lower-and-upper-bounds-on-the-incomplete-gamma-function
  [4]: https://arxiv.org/abs/2005.06384
  [5]: https://i.sstatic.net/JzNIq.png