Is it wrong to say that a Riemann sum is an unbiased estimate of an integral? Would it be wrong to say that a Riemann sum approximation of an integral
\begin{align}
    \int_a^b f(t) \mathrm{d}t \approx \sum_{k=1}^{n_\text{samples}} f(t^{\ast}_k)\Delta t, 
\end{align}
where $\Delta t = \left(b - a\right)/n_\text{samples}$, and where $t^{\ast}_k$ is the left or right end point or the midpoint of the sub-intervals is an unbiased estimate of the true integral?
The argument for Monte Carlo integral approximation with $N$ uniform samples in the interval $[a,b]$ seems to be that as the number of samples goes to infinity, then the approximation will be the exact integral with probability 1 (see e.g https://cs.dartmouth.edu/wjarosz/publications/dissertation/appendixA.pdf, page 153)
This same limit for the Riemann sum is (the definition of) the Riemann integral, hence I would argue that a Riemann sum is also unbiased.
According to a blog post (https://blog.evjang.com/2016/09/riemann-bias.html)  which I found on Google, then the Riemann sum is biased because of the deterministic steps.
But since the argument for the Monte Carlo integration being unbiased uses that $N$ goes to infinity, I can't see why the same argument can't be used for the Riemann sum approximation.
If it indeed is wrong that the Riemann sum is unbiased, I would be happy if anyone could explain the differences in the arguments.

It seems that you are swapping two different concepts here. The concepts are unbiased and consistent, which are properties of an estimator. A sequence of estimators $(T_n)_{n=1}^\infty$ is said to be unbiased for a quantity $\theta$ if, for all $n\,\in\mathbb{N}$,
$$ E[T_n] = \theta \quad.$$
It is said to be consistent if it converges in probability to $\theta$.
These are different concepts: the first says that, for every finite sample size, the average of your estimator is $\theta$. The other states that, as the sample sizes grow, the estimator getting arbitrarily close to $\theta$ with increasing probability.
Let $I = \int_a^bf(x)dx$ be your quantity of interest (assume it exists). What the most basic Monte Carlo method does is to observe that
$$I = \int_a^bf(x)dx = (b-a)\int_a^bf(x)\frac{1}{b-a}dx = (b-a)E[f(X)] \quad.$$
In the last line, we wrote the integral as being the expectation of $f(X)$, where $X$ has a uniform distribution in $(a,b)$. Hence, if we sample i.i.d. random variables $(X_i)_{i=1}^n$ with $X_1 \sim U((a,b))$, then the estimator
$$T_n = \frac{(b-a)}{n}\sum_{i=1}^nf(X_i) \quad,$$
is easily shown to be unbiased for $I$.
When you think of Riemann sums, it is usual to take a deterministic partition. If it is deterministic, then the expected value for any fixed sample size is the value of the summation it self, which in general is not the value of the integral.
