Concatenated Deletion & Sticky Channel Capacity
The identical and identically distributed (IID) binary deletion and sticky channels are both general repeat channels each characterized by a discrete probability distribution of repeating an input bit in the output $D$ times with $D \in 0, 1, 2, 3, 4, \dots$ with $D\!=\!0$ meaning that the input bit does not appear in the output of the channel at all. The question's probabilities for the deletion channel (first channel, hence the subscript $1$) are given by:

$$P(D_1) = \left[\tfrac{1}{2}, \tfrac{1}{2}, 0, 0, 0, \dots\right],\tag{1}$$

which is to say $P(D_1\!=\!0)=P(D_1\!=\!1)=\frac{1}{2},$ as given in the question.

An IID sticky channel with input duplication probability $\frac{1}{2},$ as given in the question, and of a variety that may at most double the number of bits, is also a general repeat channel, with $D_2\text{-}$times repeat probabilities:

$$P(D_2) = \left[0, \tfrac{1}{2}, \tfrac{1}{2}, 0, 0, \dots\right]\tag{2}$$

Generally, without hard-coding the above numbers, the concatenation (cascade, in series) of the above two channels into a single channel is also a general repeat channel with $D_1D_2\text{-}$times repeat probabilities:

$$P(D_1D_2) = \left[P(D_1\!=\!0) + P(D_2\!=\!0)\\- P(D_1\!=\!0)P(D_2\!=\!0),\\
P(D_1\!=\!1)P(D_2\!=\!1),\\
P(D_1\!=\!2)P(D_2\!=\!1) + P(D_1\!=\!1)P(D_2\!=\!2),\\
P(D_1\!=\!3)P(D_2\!=\!1) + P(D_1\!=\!1)P(D_2\!=\!3),\\
P(D_1\!=\!4)P(D_2\!=\!1)\!+\!P(D_1\!=\!1)P(D_2\!=\!4)\!\\+\!P(D_1\!=\!2)P(D_2\!=\!2),\\
\dots\right]\tag{3}$$

These are probabilities for the possible values of the product of independent variables $D_1$ and $D_2.$ For example the last entry before the ellipsis collects probabilities for the three mutually exclusive events $4\times1=$ $1\times4=$ $2\times2$ $=4.$
Note that $P(D_1D_2) = P(D_2D_1),$ so it does not matter which channel comes first. Using the values of $P(D_1)$ and $P(D_2)$ from Eqs. 1 &amp; 2, we get:

$$P(D_1D_2) = \left[\tfrac{1}{2}, \tfrac{1}{4}, \tfrac{1}{4}, 0, 0, \dots\right]\tag{4}$$

This characterization of the concatenated channel appears more complex than the same for the deletion and sticky channels. For each of the latter, finding the capacity appears to be an open research problem (now at the end of 2018). So I doubt the capacity of the concatenated channel is known either. However, it may be possible to find upper and/or lower bounds. For the capacity of the deletion channel, there seems to be a lower bound $1/18,$ see: Capacity of a Binary Deletion Channel.

Your question says that there should be no error correction. As far as I understand, the channel capacity does not depend on error correction, but if a working code can be demonstrated, then it gives a lower bound for channel capacity.

Mahdi Cheraghchi calls the general repeat channels $\mathcal{D}\text{-}$channels, where $\mathcal{D}$ would be the distribution of $D=D_1D_2$ for the concatenated channel. Perhaps you will find something of interest in their work (Cheraghchi 2018):


  A $\mathcal{D}\text{-}$repeat channel, given a bit,
  draws an independent sample $D ≥ 0$ from $\mathcal{D}$
  and outputs $D$ copies of the received bit.





  Our framework can be applied to obtain capacity upper
  bounds for any repetition distribution (the deletion and
  Poisson-repeat channels corresponding to the special cases
  of Bernoulli and Poisson distributions). 


References


Mahdi Cheraghchi. 2018. Capacity upper bounds for deletion-type channels. In Proceedings of the 50th Annual ACM SIGACT Symposium on Theory of Computing (STOC 2018). ACM, New D_2ork, ND_2, USA, 493-506. DOI: https://doi.org/10.1145/3188745.3188768 (arD_1iv:1711.01630 [cs.IT])