Is $\mathbb Z[x]/(x-2)=\mathbb Z$?
If we consider the evaluation map $\mathbb Z[x] \to \mathbb Z$ sending $x \rightsquigarrow 2$, then I believe the kernel of the map is $(x-2)$? I am not sure about this since not every ideal of $\mathbb Z[x]$ is principal.
If this is true then we know that $\mathbb Z[x]/(x-2) = \mathbb Z[2]$. But since $2 \in \mathbb Z$, then $\mathbb Z[2]=\mathbb Z$. Is this last implication correct?
Answering the original question first. Yes it is true. The reason is, taking any polynomial $p(x)=\sum a_ix^i$. Then we know that $x-2\equiv0$ which is equivalent to "when we see $x$, we transform it into $2$". So the polynomial above is congruent to $\sum a_i(2)^i$ which is an integer. So every polynomial can be transformed into integers, and (of course) all integers are inside the quotient, and is pairwisely not congruent to each other (since $n\equiv 0\implies n=0$). So it finishes the proof.
Your last implication is correct. We can see that we have essentially substituted $x$ as $2$ with the equivalence relationship. An extra mind-teaser is to find what $\mathbb{Z}[x]/(2x-1)$ means. I had an assignment problem on this and its structure kind of nice, so that I put it here for your reference too.