UPDATE 20220330: New answer following question clarification; old answer retained to make sense of the comments.
I think that what you are asking is whether the bits of $y$ act as a hard-core function on the inverse of a one-way function (in this case the discrete logarithm function modulo $n$). For background on hard-core functions see for example section 2.4 for Foundations of Cryptography). However, if the inverse of a one-way function is easy to compute (which is true in your case as the exponentiation function can be computed in polynomial time), then there are no hard-core functions.
Cryptographers don't phrase this in terms of uniform distribution, but in terms of discriminators that can be computed in polynomial time and offer non-trivial advantage (see definition 2.4 of the notes). They say that a predicate $b(y)$ is hard-core for $f$ if for all polynomial time discriminators we have
$$\mathbb P(A(f(U_n)),1^n)=b(U_n)<1/2+1/p(n).$$
In your case $f$ is the function $y=g^x\mod n\mapsto x$ and your function $b$ is the $i$th bit of $y=g^x\mod n$. However, I have the counterexample discriminator $A(z,1^n)$ which is to compute $g^z\mod n$ (in polynomial time) and look at the $i$th bit. This discriminates answers with probability 1 because with first argument $f(y)=x$ it returns $b(y)$.
In other words there is a computationally verifiable lack of uniformity because I can quickly test $x$ values to see whether or not they produce output that lies in $Y'$.
Old answer.
Yes. Let $|Y'|=M$ and let $z$ be any element of $Y'$ then Bayes' theorem tells us that
$$\mathbb P(g^x\mod n=z|g^x\mod n\in Y')=\frac{\mathbb P(g^x\mod n=z)\mathbb P(g^x\mod n\in Y'|g^c\mod n=z)}{\mathbb P(g^x\mod n\in Y')}.$$
We now note that $\mathbb P(g^x\mod n=z)=1/\phi(n)$ (by the uniformity noted in the question), $\mathbb P(g^x\mod n\in Y'|g^c\mod n=z)=1$ and that $\mathbb P(g^x\mod n\in Y')=M/\phi(n)$ (again by the uniformity in the question). Thus
$$\mathbb P(g^x\mod n=z|g^x\mod n\in Y')=1/M$$
for all $z\in Y'$ which describes a uniform distribution.
