For a standard Poisson process, this won't be possible. (See this question and its answer.)
Edit: Given the comments perhaps I should provide more detail.
With probability one, for every pair $0 < p < q$, $p,q$ rationals, $X(q)−X(p)$ is a non-negative integer. Since X is cadlag the same property must hold for every real pair $0 < s < t$, i.e. $X$ is increasing and integer-valued.
Let us also show that $X$ has no jumps of size more than one: with probability one, for all $x > 0$, $X(x^-) := \lim_{y \uparrow x} X(y) \geq X(x)-1$. If this failed to hold then there would be $\epsilon > 0$ and $t < \infty$ so that $$ \mathbb{P}(\exists x \in [0,t), X(x)-X(x^-) \geq 2) > \epsilon/2. $$ But since $X$ is increasing, for any positive integer $n$ we can bound this probability from above by $$ \sum_{1 \leq i < 2n} \mathbb{P}(X((i+1)t/2n)-X((i-1)t/2n) \geq 2) $$ the point being that these intervals are chosen to overlap so that a jump of size $\geq 2$ must fall in at least one of them. Each of the differences above is distributed as Poisson$(t/n)$, so the associated probability is $o(n^{-1})$ as $n \to \infty$ and thus the whole sum tends to zero as $n \to \infty$.
We then know that a process $X$ such as you describe must be increasing and integer valued, with all jumps of size $1$. In other words, $X$ is a point process on $[0,\infty)$. Now the answer from the other thread implies that $X$ must be a rate one Poisson process.