Is the $L$ in $LU$ factorization unique?

The usual theorem is the following :

Let $A \in \mathbb{C}_n^n $ be a square matrix such that its principal minors are not equal to $0$. Then there is a unique couple $(L,U)$ such that $L$ is lower triangular and $diag(L) = (1,\cdots,1)$, $U$ is upper triangular and $A = LU.$

The uniqueness is obtained thanks to the condition $$diag(L) = (1,\cdots,1).$$

There is nothing wrong with your reasoning: you will still get an LU factorization, just a different one as compared to otherwise. LU factorizations are, as you have just discovered, not unique. Uniqueness would need some extra constraints on the form of L and U.

Wikipedia states that one such condition is to force the diagonal entries of one of the matrices to all be one (so if L has a diagonal of ones, or U has, then uniqueness holds). See the link for further information: https://en.wikipedia.org/wiki/LU_decomposition#Existence_and_uniqueness

Just to add one observation to the other (good) responses. The reason why there is no unique pair of lower/triangular matrices L and U such that A=LU, is simply because the problem is underdetermined. If you count the number entries to determine in L and U you obtain $n^2+n$, while the equality A=LU only gives you $n^2$ equations that the coefficients of L and U must fulfill.

Therefore, in order to hope for uniqueness, you must add some constraints. Typically, we require the diagonal entries of L to be 1's (notice that this is completely arbitrary, we may as well require the diagonal entries of U to be $1,2,3,\ldots,n$ for what it matters). As others pointed out, this constraint is not in general sufficient to even be sure that a factorization exists. However, under certain conditions (such as those in the theorem pointed out by Dubussy), such a factorization exists and is unique.