If $L$ and $L'$ are any two left inverses of $A$, then, if $LB=L'B$, then $X=LB$ solves $AX=B.$ I had this thought reading this solution (I was advised to ask this question as a new post, so here I asked.):

If $L$ and $L'$ are any two left inverses of $A$, then, if $LB=L'B$, then $X=LB$ solves $AX=B.$

Note that $X=LB$ may not be a solution to $AX=B$ if there is no $X$ that solves $AX=B$, see example in quoted link. But if there is a solution to $AX=B$ and $A$ has a left inverse $L$, then that solution is unique and $X=LB$ is only solution to $AX=B$. 
What I am claiming is if $LB=L'B$ for any left inverses $L$ and $L'$ of $A$, then $AX=B$ has a solution: Is this correct?
For example, if $A=\begin{bmatrix}1&0\\0&1\\0&0 \end{bmatrix}$ and $B=\begin{bmatrix}3\\2\\0 \end{bmatrix}$ then we see that $L=\begin{bmatrix}1&0&0\\0&1&0\\ \end{bmatrix}$ and $L'=\begin{bmatrix}1&0&1\\0&1&1\\ \end{bmatrix}$ are two left inverses of $A$ and $LB=L'B=\begin{bmatrix}3\\2 \end{bmatrix}$ which solves $AX=B$. Does this assertion hold generally?
 A: Your claim is true.
Without loss of generality we may assume that $B$ is a column vector since we may apply the argument to each column of the matrix equation $AX=B$. So henceforth let us write $\mathbf{b}$ for $B$.
First, let $L$ and $L'$ be any two left-inverses. Then $(L'-L)A = 0$ so the set of all left-inverses is parametrized by $L+N$, where $N$ is any appropriately sized matrix such that $NA=0$.
The condition $L\mathbf{b} = L'\mathbf{b}$ for all left-inverses of $A$ therefore translates to the statement that $N\mathbf{b} = \mathbf{0}$ for all $N$ such that $NA = 0$. The key now is the following lemma.
Lemma: Let $A$ be an $m \times n$ matrix. Let $S$ denote the set of all $n\times m$ matrices $N$ such that $NA = 0$. Then
$$\bigcap_{N\in S} \ker(N) = \mathrm{col}(A).$$
Proof: Clearly the column space of $A$ is contained within the kernel of each $N\in S$, so we must have
$$\mathrm{col}(A) \subseteq \bigcap_{N\in S}\ker(N).$$
On the other-hand, let $\mathbf{v} \in \mathbb{R}^m$ be an arbitrary vector which is not in $\mathrm{col}(A)$. Then there must exist some $\mathbf{w} \in \mathrm{col}(A)^\perp$ such that $\mathbf{w}\cdot \mathbf{v} \neq 0$, for otherwise we would have $\mathbf{v} \in \left(\mathrm{col}(A)^\perp\right)^\perp = \mathrm{col}(A)$, contrary to assumptions. 
Take $\tilde{N}$ to be the matrix where every row is $\mathbf{w}$. Then we have $\tilde{N}\in S$ and $\mathbf{v} \notin \ker(\tilde{N})$. This shows that for every $\mathbf{v} \notin \mathrm{col}(A)$, there exists some $\tilde{N} \in S$ such that $\mathbf{v} \notin \ker(\tilde{N})$. Consequently, $\bigcap_{N\in S} \ker (N)$ cannot contain any vectors not in $\mathrm{col}(A)$. This proves the lemma. $\square$
Now, by our lemma above, the fact that $N\mathbf{b} = \mathbf{0}$ for all $N \in S$ implies that $\mathbf{b} \in \mathrm{col}(A)$ and hence the equation
$$A\mathbf{x} = \mathbf{b}$$
is consistent. Note that this argument is more general than requiring equality of all left-inverses. The lemma we proved above doesn't have any conditions on the size or rank of $A$. What we have shown is that 
$$AX = B$$
is satisfiable if and only if $NB = 0$ for all $N$ such that $NA = 0$.
