Cool problem. The process you are after is certainly not unique, here is a reasonably explicit construction of an increasing jump process $Z_t$ with the property you want.(U nder a couple of assumptions which I think are implicit in your statement).
The assumptions: 1) $X$ is a positive random variable, $P(X>0)=1$. 2) $X$ is unbounded, $P(X>T)\neq 0$. (In fact the construction works without the second assumption, but then $Z_t$ stops after some random time.)
Let $X_0, X_1,X_2,\ldots$ be the following Markov chain:
1.) $X_0=0$ with probability one.
2.) Given $X_0,\ldots,X_{j-1}$ let the distribution of $X_j$ be
$$P(X_j >T|X_1,\ldots,X_{j-1}) = P(X >T|X>X_{j-1}).$$
(In particular $X_1$ is a "copy" of $X$: $P(X_1 >T)=P(X>T)$. The distributions of $X_2,\ldots$ are more complicated.)
Clearly $X_j$ is a strictly increasing sequence with probability one. One can also show that $\lim_n X_n =\infty$ with probability one. Since $X_0=0$ it follows that for any $t\ge 0$ we can find a unique (random) $j_t$ such that
$$X_{j_t-1} \le t< X_{j_t}.$$
Let
$$Z_t = X_{j_t},$$
so $Z_t$ is piecewise constant and increasing.
To see that $Z_t$ has the property you want, note that
$$P(Z_t >T)=\sum_{j=1}^\infty P( (X_j>T) \& (j_t =j)).$$
Let $\nu$ be the probability measure for the distribution of $X_{j-1}$. Then by the definitions of $j_t$ and of $X_j$,
$$P( (X_j>T) \& (j_t =j)) = \int_{(0,t]} P((X >T) \& (X>t) |X>x) d \nu(x).$$
Since the even $(X >t) \subset (X>x)$ for $x \le t$ in the domain of integration we have
$$P((X>T) \& (X>t) |X>x) = \frac{P((X>T)\& (X>t))}{P(X>t)} \frac{P(X>t)}{P(X>x)} =P(X>T|X>t) P(X>t|X>x).$$
Thus
$$P( (X_j>T) \& (j_t =j)) = P(X>T|X>t) P((X_j >t) \& (X_{j-1}\le t)) = P(X>T|X>t)P(j_t=j),$$ and so $$P(Z_t >T) = P(X>T | X>t) \sum_{j=1}^\infty P(j_t=j)= P(X>T | X>t)$$
as desired!
(By the way, you can construct the sequence $X_1,\ldots$ as follows. Let $Y_1,\ldots$ be a sequence of independent identically distributed random variables, each with the distribution of $X$. We will take $X_j=Y_{n_j}$ with $n_j$ a certain random sequence that depends on $Y_1,\ldots$. Let $X_1=Y_1$ and given $X_j=Y_{n_j}$ let $n_{j+1}$ be the first index $n$ larger than $n_j$ such that $Y_{n_j} < Y_{n}$. So long as $P(X>T)\neq 0$ for all $T>0$ we produce in this way an infinite sequence $X_1,\ldots$.)