The answer is both yes, and no. Or, to put it another way, the answer depends on what exactly you mean by "represent probabilities", and there is a valid sense in which the answer is yes, and another valid sense in which the answer is no.
No, they don't represent the probability
No, they do not represent the true probability.
You can think of a neural network as a function $f$. Let $f(y;x)$ denote the softmax output of the neural network corresponding to class $y$, on input $x$. Then $f(y;x)$ will typically not be equal to $p(y|x)$, the probability that sample $x$ is from class $y$. $f(y;x)$ can be viewed as an estimate of $p(y|x)$ -- as a best-effort guess at $p(y|x)$ -- but it can be an arbitrarily bad estimate/guess. Neural networks routinely make errors on tasks that even humans find clear. Also, neural networks have systematic biases. For instance, as the other answer explains, neural networks tend to be biased towards "overconfidence".
So you should not assume that the output from the neural network represents the true probability $p(y|x)$. There is some underlying probability. We might not know how to compute it, but it exists. Neural networks are an attempt to estimate it, but it is a highly imperfect estimate.
Yes, they do represent probabilities
While the softmax outputs are not the true probability $p(y|x)$, they do represent a probability distribution. You can think of them as an estimate of $p(y|x)$. For a number of reasons, it is an imperfect and flawed estimate, but it is an estimate nonetheless. (Even bad or noisy estimates are still estimates.)
Moreover, the way we train neural networks is designed to try to make them a good estimate -- or as good as possible. We train a neural network to minimize the expected loss. The expected loss is defined as
$$L = \mathbb{E}_x[H(p(y|x),f(y;x))],$$
where the expectation is with respect to $x$ chosen according to the data distribution embodied in the training set, and $H$ is the cross-entropy of the distribution $f(y;x)$ relative to the distribution $p(y|x)$. Intuitively, the smaller the training loss is, the closer that $f(y;x)$ is to $p(y|x)$.
So, neural networks are trained in a way that tries to make its output be as good an approximation to $p(y|x)$ as is possible, given the limitations of neural networks and given the training data that is available. As highlighted above, this is highly imperfect. But $f(y;x)$ does still represent a probability distribution, that is our attempt to estimate the true probability distribution $p(y|x)$.