I think the criterion of having an observation exceed $\bar X + 2S$ is intended as a way to highlight an observation as an "outlier." This criterion does not work very well because the large observation 120 itself leads to inflating both $\bar X$ and $S$ so that $\bar X + 2S$ is larger than it "should" be.

A better criterion might be to delete 120 before computing mean $\bar X^\prime$ and $S^\prime.$ Then you'd have $120 > 6.082.$ [Computation using R.]

x.d = c(2,3,4,5);  mean(x.d) + 2*sd(x.d)
[1] 6.081989

Another method is to use the interquartile range (IQR) as a 'robust' measure of variability (which does not depend so crucially on any one observation). That is the idea behind using boxplots to suggest possible outliers.

x = c(2, 3, 4, 120, 5)
boxplot(x, horizontal=T)

The boxplot 'outlier' rule is to suggest any observation above $Q_3 + 1.5\text{IQR}$ as a high outlier, where $Q_3$ is the third quartile. Similarly, any observation below $Q_1 - 1.5\text{IQR}$ is suggested as a possible low outlier.

All outlier criteria need to be used with great caution because some distributions routinely have 'outliers'--especially in large samples. Almost all samples of size 100 from an exponential distribution have 'outliers'. More specifically, for the simulation below about 99% of exponential samples of size 100 had at least one outlier and the average number of outliers per sample was nearly 5.

nr.out = replicate(10^5, length(boxplot.stats(rexp(100))$out))
mean(nr.out)
[1] 4.84265
mean(nr.out > 0)
[1] 0.99023

Even among samples of size 100 from a normal distribution, over 90% have at least one 'boxplot outlier'--but usually only one or two of them per sample. Generating such a sample with today's date as seed, I happened to get three 'outliers'.

set.seed(1116); x = rnorm(100)
boxplot(x, horizontal=T)

The short answer, is no, it is not an error.

As @whuber notes, there is nothing surprising (at least to a statistician) about the fact that two standard deviations below the mean of a count variable could be a negative value. Thus, to answer your question, perhaps it would be more useful to ponder why you might find the result surprising.

Why you might be surprised

• Many introductory statistics textbooks show how you can use the mean, standard deviation, and the normal distribution to make claims like approximately 2.5% of the sample is expected to score below two standard deviations below the mean. You may have generalised this idea to a variable where the assumptions of such a procedure are invalid.
• If you did this, you would be saying to yourself: "this is strange, how is it possible for 2.5% of the data to have counts below -0.6".

Estimating percentiles for counts

• Your variable is not normally distributed, it is a count variable. It is discrete; it is a non-negative integer. Thus, in order to estimate the percentage that is greater than or equal to a given value, you need an approach suited to counts. A basic approach would involve using the sample data to estimate such percentiles. More sophisticated approaches could involve developing a model of the distribution suited to counts, justified by the data and knowledge of the phenomena, and estimated using the sample data.