So, here's my attempt. I arrive at something close to yours (but no proof), although I believe your definition of hazard is not quite correct (hazard is not a momentary risk among the survived, it's the instantaneous rate among the survived):

Hazard: $$h(t) = \lim_{\partial t \to 0} \frac {P(t ≤ T ≤ t + \partial t \,|\, T>t)} {\partial t} = \lim_{\partial t \to 0} \frac {P(t ≤ T ≤ t + \partial t)} {{\partial t} \,\cdot\, P(T>t) }$$

So using this along with $X\sim Exp(\lambda)$, $Y$ has positive support and a strictly increasing hazard, and $Z = X + Y$.

For clarity, I'll rename the timepoints, so that $Z$ occurs at time $t_Z$ and $X$ occurs at time $t$ (since this is the variable one when $t_Z$ is known).

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$$ h_{X|Z=t_Z}(t) = \lim_{\partial t \to 0} \frac {P(t≤X≤t+{\partial t} \,| \, Z=t_Z)} {{\partial t} \,\cdot\, P(X>t \,|\, Z=t_Z)} \;\;\;\; \bf{\scriptsize{(Eq. 0)}}$$

Using Bayes' theorem, the numerator and denominator can be rewritten:

Numerator: $$ P(t \leq X \leq t+{\partial t} \,| \, Z=t_Z) = \frac {P(Z=t_Z \,| \, t \leq X \leq t+{\partial t}) \,\cdot\, P(t \leq X \leq t+{\partial t})} {P(Z=t_Z)} \;\;\;\; \bf{\scriptsize{(Eq. 1)}}$$

P from Denominator: $$ P(X>t \,|\, Z=t_Z) = \frac {P(Z=t_Z \,|\, X>t) \,\cdot\, P(X>t)} {P(Z=t_Z)} \;\;\;\; \bf{\scriptsize{(Eq. 2)}}$$

Replacing these back, we get:

$$ h_{X|Z=t_Z}(t) = \lim_{\partial t \to 0} \frac {P(Z=t_Z \,| \, t \leq X \leq t+{\partial t}) \,\cdot\, P(t \leq X \leq t+{\partial t})} {{\partial t} \,\cdot\, P(Z=t_Z \,|\, X>t) \,\cdot\, P(X>t)} $$

$$ = \lim_{\partial t \to 0} \frac {P(Z=t_Z \,| \, t \leq X \leq t+{\partial t})} {P(Z=t_Z \,|\, X>t)} \cdot \frac {P(t \leq X \leq t+{\partial t})} {{\partial t} \,\cdot\, P(X>t)} $$

$$ = \frac {P(Z=t_Z \,| \, X=t)} {P(Z=t_Z \,|\, X>t)} \cdot h_X(t) \;\;\;\; \bf{\scriptsize{(Eq. 3)}} $$

This can then be rewritten in the form you'd presented:

$$ h_{X|Z=t_Z}(t) = \frac {P(Y=t_Z - t)} {P(Z=t_Z \,|\, X>t)} \cdot \lambda \;\;\;\; \bf{\scriptsize{(Eq. 4)}} $$

Note, this is almost identical to what you found, except $X>t$ rather than $X \geq t$.

It is not immediately obvious to me that this must be decreasing. You don't strictly need to show that the denominator is increasing and that the numerator is decreasing (that's far too strict). But you can try to show that the entire expression is decreasing (e.g., by quotient rule).

I currently haven't gotten anywhere with that, and maybe you'd need a distribution for $Y$ to do it.

Please let me know if I've made any errors.

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$\bf{UPDATE:}$

Tried a different way. Rather than starting by estimating the conditional hazard of X, let's just look at the distribution of X:

$$ P(X=t|Z=t_Z) = \frac {P(Z=t_Z|X=t) \cdot P(X=t)} {P(Z=t_Z)} $$

$$ = \frac {P(Y = t_z-t) \cdot P(X=t)} {P(Z=t_Z)} $$

$$ = \frac {P(Y = t_z-t) \cdot P(X=t)} { \int_{s=0}^{t_Z} P(X=s) \cdot P(Y=t_Z-s) \,ds } \;\;\;\; \bf{\scriptsize{(Eq. B.1)}} $$

Observe here that the denominator is independent of $t$ (and so a constant). The numerator is a product of a distribution with constant hazard ($X$) and one with monotonically increasing hazard ($Y$). If we take Y to follow a Weibull distribution with $k > 1$, the two variables have:

$$ P(X=t) = \lambda_E e^{-\lambda_E t} \;\;\;\; \bf{\scriptsize{(Eq. B.2)}} $$

and

$$ P(Y=t_z-t)= \frac{k}{\lambda_W} (\frac{t_z-t}{\lambda_W})^{k-1} e^{-(\frac{t_z-t}{\lambda_W})^k} \;\;\;\; \bf{\scriptsize{(Eq. B.3)}} $$

Where the $\lambda$s from the exponential and Weibull distribution are denoted $\lambda_E$ and $\lambda_W$, respectively. We can also denote the denominator from Eq. B.1 as follows:

$$ \theta = \int_{s=0}^{t_Z} P(X=s) \cdot P(Y=t_Z-s) \,ds \;\;\;\; \bf{\scriptsize{(Eq. B.4)}} $$

Combining all of these, we get:

$$ P(X=t|Z=t_Z) = \frac {1} {\theta} \frac {k} {\lambda_W} (\frac{t_Z-t}{\lambda_W})^{k-1} e^{-(\frac{t_Z-t}{\lambda_W})^k} \cdot \lambda_E e^{- \lambda_E t} $$

$$ = \frac{\lambda_E \cdot k}{\theta \cdot \lambda_W^{k} } \cdot \ (t_Z-t)^{k-1} \cdot\ e^{- ( \frac {t_Z-t} {\lambda_W})^k - \lambda_E t} \;\;\;\; \bf{\scriptsize{(Eq. B.5)}} $$

This is a work in progress, and I'll get back to it later to see if I can evaluate the hazard function associated with it. My immediate idea is to simplify it with the Weibull dist that has $k=2$ and perhaps show that the hazard is actually increasing in that case. Update pending.

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$\bf{UPDATE \;2:}$

Stuck currently. However, ran some simulations with Weibull, and the conditional hazard of X seems to be increasing when the hazard of Y is increasing (Weibull has a shape parameter > 1), although this is also the case for some shape parameters below 1 (down to ~0.4).

library(rstpm2)
library(ggplot2)

n <- 10000
Y <- rweibull(n, shape = 2, scale = 1)
X <- rexp(n, rate = 1)
Z <- X + Y

df <- data.frame(cbind(status = rep(1,length(X)), X, Z))

fit <- stpm2(Surv(X,status==1)~Z, data=df, df=3)

hazard <- predict(fit, newdata=data.frame(Z=2),
                  type="hazard", grid=TRUE, full=TRUE, se.fit=TRUE)

ggplot(hazard,
       aes(x=X,y=Estimate,ymin=lower,ymax=upper,fill=Z)) +
  xlab("Time") +
  ylab("Hazard") +
  geom_ribbon() +
  geom_line() +
  scale_y_continuous(trans='log2')

Here, you can see the growth of the conditional hazard being less than exponential.

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$\bf{UPDATE \;3:}$

Couldn't quite let it go. The closest I'm gonna get is probably expressing the conditional hazard of $X$ purely in terms of marginal probabilites of $X$ and $Y$.

Let's once again take:

$$ h_{X|Z=t_Z}(t) = \lim_{\partial t \to 0} \frac {P(t≤X≤t+{\partial t} \,| \, Z=t_Z)} {{\partial t} \,\cdot\, P(X>t \,|\, Z=t_Z)} \;\;\;\; \bf{\scriptsize{(Eq. 0)}} $$

Replacing the denominator as we did in $Eq. 2$ can be extended as follows:

$$ P(X>t \,|\, Z=t_Z) = \frac {P(Z=t_Z \,|\, X>t) \,\cdot\, P(X>t)} {P(Z=t_Z)} \;\;\;\; \bf{\scriptsize{(Eq. 2)}} $$

$$ = \frac {P(Z=t_Z-t \,|\, X>0) \,\cdot\, P(X>t)} {P(Z=t_Z)} $$

$$ = \frac {P(Y=t_Z-t-X) \,\cdot\, P(X>t)} {P(Z=t_Z)} $$

$$ = \frac {P(Y=t_Z-t-X) \,\cdot\, P(X>t)} {P(Z=t_Z)} $$

$$ = \frac { (\int_{s=0}^{t_Z-t} P(X=s) \cdot P(Y=t_Z-t-s) \,ds) \,\cdot\, P(X>t)} {P(Z=t_Z)} \;\;\;\; \bf{\scriptsize{(Eq. C.1)}} $$

And the numerator as we did in $Eq. 1$:

$$ P(t \leq X \leq t+{\partial t} \,| \, Z=t_Z) = \frac {P(Z=t_Z \,| \, t \leq X \leq t+{\partial t}) \,\cdot\, P(t \leq X \leq t+{\partial t})} {P(Z=t_Z)} \;\;\;\; \bf{\scriptsize{(Eq. 1)}}$$

Leaves:

$$ h_{X|Z=t_Z}(t) = \lim_{\partial t \to 0} \frac {P(Z=t_Z \,| \, t \leq X \leq t+{\partial t}) \,\cdot\, P(t \leq X \leq t+{\partial t})} {{\partial t} \,\cdot\, (\int_{s=0}^{t_Z-t} P(X=s) \cdot P(Y=t_Z-t-s) \,ds) \,\cdot\, P(X>t)} $$

$$ = \lim_{\partial t \to 0} \frac {P(t \leq X \leq t+{\partial t})} {{\partial t} \,\cdot\, P(X>t)} \cdot \frac {P(Z=t_Z \,| \, t \leq X \leq t+{\partial t})} {\int_{s=0}^{t_Z-t} P(X=s) \cdot P(Y=t_Z-t-s) \,ds} $$

$$ = h_X \cdot \frac {P(Z=t_Z \,| \, X=t)} {\int_{s=0}^{t_Z-t} P(X=s) \cdot P(Y=t_Z-t-s) \,ds} $$

$$ = \lambda \cdot \frac {P(Y=t_Z-t)} {\int_{s=0}^{t_Z-t} P(X=s) \cdot P(Y=t_Z-t-s) \,ds} \;\;\;\; \bf{\scriptsize{(Eq. C.2)}}$$

I hope I haven't made any errors, because this feels like it's getting closer to something workable.