Finding a maximal independent set in parallel
There is a minor mistake, the probability of $v$ being minimal is $1/(d+1)$. That doesn't change anything, but it is still worth pointing out.

The problem is that the events you are summing aren't disjoint. Consider two vertices that aren't adjacent but have a common neighbour. If both vertices end up being minimal amongst their neighbours, then the common neighbour gets counted as being coloured twice.

We need to examine what happens in a vertex more closely. Let's compute the probability (and thus expectation of the indicator variabele) of a single vertex getting coloured.

Assume for the sake of analysis that the graph is $d$-regular and contains no triangles or squares. We will compute the chance that a given vertex $v$ does not get coloured. There are $d+1$ possibilities: the $v$ might be the smallest amongst it neighbours, $v$ might be the second smallest, third smallest, and so on... We are not interested in the case that it is the smallest, since then it gets coloured.

If $v$ is the second smallest, there is one vertex that is smaller. The probability of this vertex being smallest amongst its neighbours is $\frac{1}{d}$ (and thus $v$ getting coloured). The probability of the remaining neighbours being smallest amongst their neighbours is 0 (since $v$ is smaller). The total probability (of $v$ not getting coloured) from this case is $\frac{1}{d+1}\cdot\frac{d-1}{d}$.

If $v$ is third smallest, there are two smaller vertices. The probability from this case is $\frac{1}{d+1}\cdot(\frac{d-1}{d})^2$.

Continuing on like this, we find that the probability that $v$ does not get coloured is $\frac{1}{d+1}\Sigma_{i=1}^d (\frac{d-1}{d})^i$. The probability of a given vertex getting coloured is thus $1-\frac{1}{d+1}\Sigma_{i=1}^d (\frac{d-1}{d})^i$. As $d\to \infty$, the probability becomes $\frac{1}{e}\approx 0.368$.

Hence the probability that a given vertex gets coloured is a constant, so the expected number of vertices that gets coloured in the first step is indeed $O(n)$.

That does not make the algorithm $O(1)$. Assuming the probability stays constant (which is not entirely justified considering nodes get coloured and no longer participate in the algorithm, the graph does not remain $d$-regular), in the $k$-th step there will be (in expectation) ${(\frac{1}{e})}^kn$ uncoloured nodes. The decay is exponential, so the algorithm takes $O(\log n)$ steps.

Perhaps somebody else will be able to offer a more precise analysis, but from my argumentation it seems likely the algorithm is $O(\log n)$.