As I understand from a comment, the OP didn't realize that the Kaplan-Meier estimate is nothing but the empirical estimate of the survival function in case when there is no censoring.
Let me tell a word about that. Consider two independent random variables $X$ and $Y$ with continuous distributions, and independent replicated observations $x_i$ and $y_i$, $i=1, \ldots, n$. In the context of the Kaplan-Meier estimate, $Y$ is considered as the censoring variable and one observes the minima $t_i=\min(x_i,y_i)$ together with the indicators $\delta_i={\boldsymbol 1}_{x_i \leq y_i}$, independent replicated observations of $T=\min(X,Y)$ and $\Delta={\boldsymbol 1}_{X \leq Y}$ respectively.
Note that $\Pr(T >t)=\Pr(X>t)\Pr(Y>t)$, that is to say $\boxed{S^T(t)=S^X(t)S^Y(t)}$ by denoting $S^T$, $S^X$ and $S^Y$ the survival functions of $T$, $X$ and $Y$ respectively.
The usual empirical survival function $\hat{S}^T$ of $T$ is available from the data. When seeking estimates $\hat{S}^X$ and $\hat{S}^Y$ of $S^X$ and $S^Y$, it is natural to require the empirical analogous of the previous boxed formula, that is to say $\boxed{\hat{S}^T(t)=\hat{S}^X(t)\hat{S}^Y(t)}$.
Then remember that:
The Kaplan-Meier estimates of $S^X$ and $S^Y$ satisfy this relation (at least when there are no ties, I don't know and I have not checked when there are ties). The case when $Y=+\infty$ corresponds to the absence of censoring, in this case $T=X$, $S^Y\equiv 1$, $\hat{S}^Y\equiv 1$ and one gets $\hat{S}^T(t)=\hat{S}^X(t)$: the Kaplan-Meier estimate is nothing but the empirical estimate of the survival function.
In fact (at least when there are no ties), the Kaplan-meier estimates can even be derived from the required relation $\boxed{\hat{S}^T(t)=\hat{S}^X(t)\hat{S}^Y(t)}$, after requiring in addition that $\hat{S}^X$ and $\hat{S}^Y$ are step functions jumping at the observations of $x_i$ ($t_i$ when $\delta_i=1$) and $y_i$ ($t_i$ when $\delta_i=0$) respectively.
