We see here a proof of the random-time memoryless property $P(X>T+s|X>T)=P(X>s)$ where $E\sim Exp(\lambda)$ and $T\ge 0$ is a continuous random variable independent of $E$.
The proof, however, relies upon the existence of a pdf $f_T$ for $T$ (and hence the continuity of $T$); this places restrictions on the random variable $T$ that I'd like to avoid. Can we prove this result independent of such a pdf for $T$? I imagine conditioning on $E$ instead should be possible in deriving a proof, but can't seem to find anything on this.
You do not require $T$ to be continuously distributed. It is only required to be almost surely non-negative and independent from $X$.
For instance: Let $T$ be a discrete non-negative integer-valued random variable ($T\in\Bbb N$).
$$\begin{align}\mathsf P(X\gt T+s\mid X\gt T) &=\dfrac{\mathsf P(X\gt T+s)}{\mathsf P(X\gt T)}\\[1ex]&=\dfrac{\displaystyle\sum_{t\in\Bbb N}\mathsf P(X\gt t+s\mid X\gt t)\,\mathsf P(T=t)}{\displaystyle\sum_{t\in\Bbb N}\mathsf P(X\gt t\mid X\gt t)\,\mathsf P(T=t)}\\[1ex]&~~\vdots\\[1ex]&=\mathsf P(X\geqslant s)\end{align}$$
So more generally.
$$\begin{align}\mathsf P(X\gt T+s\mid X\gt T) &=\dfrac{\mathsf P(X\gt T+s)}{\mathsf P(X\gt T)}\\[1ex]&=\dfrac{\mathsf E(\mathsf E(\mathbf 1_{X\gt T+s}\mid T))}{\mathsf E(\mathsf E(\mathbf 1_{X\gt T}\mid T))}\\[1ex]&=\dfrac{\mathsf E(\mathrm e^{-\lambda(T+s)})}{\mathsf E(\mathrm e^{-\lambda(T)})}\\[1ex] &= \mathsf e^{-\lambda s}\\[1ex]&=\mathsf P(X\gt s)\end{align}$$