Let $X$ be exponentially distributed with mean $1$ and $q \in (0,1)$. Define the random variable $Y \triangleq (1-q)X + q$. Now, the CCDF of Y is given by $\mathbb{P}(Y>y) = e^{-\frac{y-q}{1-q}}\mathbf{1}(y\geq q) +\mathbf{1}(y<q)$, where $\mathbf{1}(\cdot)$ is the indicator function. Let $Z>0$ be a random variable. I claim that the following statements are both valid: $$ (1) \hspace{0.5cm} \mathbb{P}(Y> Z) = {\mathbb{E}} \left[ e^{-\frac{Z-q}{1-q}}\mathbf{1}(Z\geq q)+\mathbf{1}(Z<q) \right]= {\mathbb{E}} \left[ e^{-\frac{Z-q}{1-q}}\mathbf{1}(Z\geq q)\right]+\mathbb{P}(Z<q), $$ $$ (2) \hspace{0.5cm} \mathbb{P}(Y> Z) = {\mathbb{E}} \left[ e^{-\frac{(Z|(Z \geq q))-q}{1-q}}\right]\mathbb{P}(Z\geq q) +\mathbb{P}(Z<q), $$ with the conditional random variable $(Z|(Z \geq q))$.
The title of the post is because combining $(1)$ and $(2)$ would follow
$$ \mathcal{L}_{(Z|(Z \geq q))}\left( \frac{1}{1-q}\right)\triangleq{\mathbb{E}} \left[ e^{-\frac{(Z|(Z \geq q))}{1-q}}\right] = \frac{{\mathbb{E}} \left[ e^{-\frac{Z}{1-q}}\mathbf{1}(Z\geq q)\right] }{\mathbb{P}(Z\geq q)}. $$
For me, both (1) and (2) are valid in their logical sense. However, I struggle to formally derive (2) from (1) or vice versa.
In view of mathematical rigor, there are two primary concerns regarding your assertions:
1. The assertion only holds if we assume $Z$ to be independent of $Y$ (and, consequently, of $X$). Take, for instance, $Z = Y$. Then the stated equalities don't hold true.
2. The term "conditional random variable" lacks a precise definition in mathematical terms. Generally, for any event $E$ with $\mathbb{P}(E) > 0$, it's not possible to "conditionalize" a random variable $Z$ (by means of processing $Z$ as a quantity) in such a way that all associated probabilistic quantities (like the distribution of $Z$ or joint distributions involving $Z$) are "conditionalized" in a compatible manner. For this reason, people do not bother to define the notion of conditional random variable.
Rather than attempting to manipulate the random variable itself, a more natural and neat solution is to replace the probability law $\mathbb{P}$ by the conditional probability law $\mathbb{P}(\cdot \mid E)$.
In light of this discussion, now we can properly adjust the assertions and prove them rigorously.
Suppose $X$ is an exponential RV under $\mathbb{P}$ with mean $1$. Fix $q \in (0, 1)$ and define $Y = (1-q)X + q$. As OP has already observed, the CCDF $\tilde{F}_Y$ of $Y$ is given by
$$ \tilde{F}_Y(y) = \mathbb{P}(Y > y) = e^{-\frac{y-q}{1-q}}\mathbf{1}(y \geq q) + \mathbf{1}(y < q). $$
Also, let $Z$ be a positive RV independent of $Y$. Then by the law of iterated expectation,
\begin{align*} \mathbb{P}(Y > Z) &= \mathbb{E}[\mathbb{P}(Y > Z \mid Z)] \\ &= \mathbb{E}[\tilde{F}_Y(Z)] \\ &= \mathbb{E}\left[ e^{-\frac{Z-q}{1-q}}\mathbf{1}(Z \geq q) + \mathbf{1}(Z < q) \right] \\ &= \mathbb{E}\left[ e^{-\frac{Z-q}{1-q}}\mathbf{1}(Z \geq q) \right] + \mathbb{P}(Z < q), \end{align*}
proving $(1)$. Moreover, if $\mathbb{P}(Z \geq q) > 0$, then by noting that $\mathbb{E}[ \square \mid Z \geq q] = \frac{\mathbb{E}[ \square \mathbf{1}(Z \geq q)]}{\mathbb{P}(Z \geq q)}$ for any placeholder RV $\square$, we get
\begin{align*} \mathbb{P}(Y > Z) &= \mathbb{E}\left[ e^{-\frac{Z-q}{1-q}} \,\middle|\, Z \geq q \right] \mathbb{P}(Z \geq q) + \mathbb{P}(Z < q). \end{align*}
Now note that the distribution of $Z$ under $\mathbb{P}(\cdot \mid Z \geq q)$ is precisely the conditional distribution $Z \mid (Z \geq q)$ under $\mathbb{P}$, part $\text{(2)}$ is also established.