Estimating $\theta$ based on censored data when $X_i\sim \text{Uniform}(0,\theta)$ with $\theta\ge 1$ Suppose $(X_i)_{1\le i\le n}$ are i.i.d $\text{Uniform}(0,\theta)$ random variables where $\theta \ge 1$. We observe $Y_i=\min(X_i,1)$ instead of $X_i$. I wish to estimate $\theta$ based on the data $(Y_i)_{1\le i\le n}$.
I think the likelihood takes the form
\begin{align}
L(\theta\mid \boldsymbol y)&=\prod_{i:x_i<1} \left(\frac1{\theta}\mathbf1_{0<y_i<\theta}\right)\prod_{i:x_i\ge 1}\left(1-\frac1{\theta}\right)
\\&=\theta^{-r}(1-\theta^{-1})^{n-r}\mathbf1_{0<y_{(n)}<\theta}\,,
\end{align}
where $r=\sum_{i=1}^n \mathbf1_{x_i<1}=\sum_{i=1}^n \mathbf1_{y_i<1}$ is the number of observations less than $1$.
If this is the correct likelihood, then a sufficient statistic for $\theta$ seems to be $T=(R,Y_{(n)})$. Or is the sufficient statistic simply $R$? Is the sufficient statistic complete? This would help to answer if there is UMVUE of $\theta$.
On the other hand, what is MLE of $\theta$? If I ignore the indicator $\mathbf1_{0<y_{(n)}<\theta}$ in the likelihood and assume $\theta\ne 1$, then differentiation leads to the stationary point $\hat\theta =\frac{n}{r}$. I am not sure if this is a valid answer.

For censoring problems like this you are dealing with an observable random variable that is a mixture of a continuous and discrete part, with the discrete part occurring at the censoring value.  For this type of data, it is sometimes easier to derive the likelihood function by starting with the CDF of the censored values and then using this to get their PDF.  I'm going to do the derivation without your assumption that $\theta \geqslant 1$ initially, and then I'll add that assumption at the end, so that you can see how it is done in general.

Deriving the PDF for the censored values: The simplest way to do this is to first obtain the CDF.  Since $X \sim \text{U}(0, \theta)$ you have:
$$\mathbb{P}(X \leqslant x) = \frac{\min(x, \theta)}{\theta}
\quad \quad \quad \text{for all } x \geqslant 0.$$
Thus, for all $y \geqslant 0$ we have the CDF:
$$\begin{align}
F_Y(y) 
&\equiv \mathbb{P}(Y \leqslant y) \\[8pt]
&= \mathbb{P}(\min(X,1) \leqslant y) \\[8pt]
&= \mathbb{P}(X \leqslant y) \cdot \mathbb{I}(y \leqslant 1) + \mathbb{I}(y > 1) \\[6pt]
&= \frac{\min(y, \theta)}{\theta} \cdot \mathbb{I}(y \leqslant 1) + \mathbb{I}(y > 1). \\[6pt]
\end{align}$$
To get the PDF for this mixture variable we use the Dirac-delta function $\delta$ for the discrete part.  Differentiating the CDF and using the Dirac delta function gives the PDF:$^\dagger$
$$\begin{align}
f_Y(y) 
&= \frac{dF_Y}{dy}(y) \\[6pt]
&= \bigg[ \frac{d}{dy} \frac{\min(y, \theta)}{\theta} \bigg] \mathbb{I}(y \leqslant 1) + \bigg[ 1 - \frac{\min(\theta, 1)}{\theta} \bigg] \delta(1) \\[6pt]
&= \bigg[ \frac{1}{\theta} \cdot \mathbb{I}(y \leqslant \theta) \bigg] \mathbb{I}(y \leqslant 1) + \frac{\theta - \min(\theta, 1)}{\theta} \cdot \delta(1) \\[6pt]
&= \frac{1}{\theta} \cdot \mathbb{I}(y \leqslant \min(\theta, 1)) + \frac{\max(0, \theta-1)}{\theta} \cdot \delta(1). \\[6pt]
\end{align}$$
Now, if you add in your assumption that $\theta \geqslant 1$ then you get the simplified PDF:
$$f_Y(y) = \frac{1}{\theta} \cdot \mathbb{I}(y \leqslant 1) + \frac{\theta-1}{\theta} \cdot \delta(1).$$

The likelihood function and MLE: Now that we have the PDF we can write the likelihood function.  Using your notation, suppose we let $R \equiv R(\mathbf{y}) \equiv \sum_{i=1}^n \mathbb{I}(y_i < 1)$ be the number of non-censored data points.  We can then write the likelihood as:
$$\begin{align}
L_\mathbf{y}(\theta)
&= \prod_{i=1}^n f_Y(y_i) \\[6pt]
&= \bigg( \frac{1}{\theta} \bigg)^r \times \bigg( \frac{\theta-1}{\theta} \bigg)^{n-r} \\[6pt]
&= \frac{(\theta-1)^{n-r}}{\theta^n}, \\[6pt]
\end{align}$$
which gives the log-likelihood function:
$$\ell_\mathbf{y}(\theta) = (n-r) \log(\theta-1) - n \log (\theta)
\quad \quad \quad \quad \quad 
\text{for } \theta \geqslant 1.$$
The statistic $R$ is sufficient for this problem, and the MLE is:
$$\hat{\theta}_\text{MLE} = \frac{n}{r}.$$
I will leave the rest of the analysis (completeness, etc.) to you.  Further analysis should be reasonably simple due to the simple form of the log-likelihood function

$^\dagger$ We use the convention that $0 \cdot \delta(x) = 0$ so that the last term disappears if $\theta < 1$.
