Robustness to non-uniform randomness vs. one-sidedness
Only one-sided deciders are robust to bad coins, under your definition; no other decider can be robust to bad coins.  Let $D$ be a decider that is not one-sided.  For $x \in \mathcal{Y}$, let $\mathcal{W}_x = \{r \in \{0,1\}^m : D(x,r)=\textsf{no}\}$ be the set of random strings $r$ such that $D$ outputs the wrong answer.  Consider a distribution $R$ that, with probability $1/3+\epsilon$, picks $r$ uniformly at random from $\mathcal{W}_x$, and with probability $2/3-\epsilon$, picks $r$ uniformly at random from its complement.  Then the decider is correct on $x$ with probability $2/3 - \epsilon$ when $r$ is chosen from $R$; yet $d_{TV}(R,U) = \epsilon$.  So, there is no $c \in [0,1)$ with the desired property; you'd need to take $c=1$.

If you relax your definition to allow $c \in [0,1]$, so that (non-one-sided) deciders robust to bad coins exist, then the answer to your question is yes, it is a strict subset.  It is not hard to come up with trivial examples.  For instance, suppose that $D$ is correct for all choices of $r$ except for a single bad value $r_\text{bad}$; if $r=r_\text{bad}$, then $D$ is incorrect.  This decider is robust to bad coins but is not a one-sided decider.

Moreover: with such an adjusted definition, every decider is robust to bad coins. By the definition of total variation distance and the definition of a decider,

$$\Pr_{r \in R}[\mathcal{W}_x] \le \Pr_{r \in U}[\mathcal{W}_x] + d_{TV}(R,U) \le 1/3 + d_{TV}(R,U),$$

so the probability that (1) holds is at least $2/3 - d_{TV}(R,U)$.  Thus, $D$ is robust to bad coins with $c = 1$.

So, I think you need to re-consider your definition of "robust to bad coins".  With your original formulation ($c \in [0,1)$), only one-sided deciders can ever meet it; with an adjusted formulation ($c \in [0,1]$), every decider meets it.  In both cases, it collapses to be equivalent to some other existing notion.  This suggests that we need a different definition.