If a matrix has multiple left inverse, then solution is not unique? I just want to clarify the statement " If A has a left inverse, Ax=b has a unique solution."
This is saying that A only has at most one left inverse correct? Because if A has more than one left inverse then x would not be unique since if Y1 and Y2 and left inverse of A, x=Y1b and Y2b are both solutions.
 A: If $A$ has a left inverse, then the linear transformation defined by $A$ is injective: because if $B$ is the left inverse, then from $Ax=Ay$ we can left multiply by $B$ to get $x=BAx=BAy = y$.
That means that $Ax=b$ has at most one solution; because if $x_0$ and $x_1$ are both solutions, then $b=Ax_0=Ax_1$, hence $x_0=x_1$.
However, the system $Ax=b$ could have no solutions. For example, the matrix
$$A=\left(\begin{array}{cc}
1 & 0\\
0 & 1\\
0 & 0
\end{array}\right)$$
has left inverses (infinitely many, in fact): namely, all matrices of the form
$$B=\left(\begin{array}{cc}
1 & 0 & a\\
0 & 1 & b
\end{array}\right)$$
with $a,b$ arbitrary are left inverses of $A$.
For any $b$ of the form $(r,s,0)^T$, the system $Ax=b$ has exactly one solution; for $b$ of the form $(r,s,t)^T$ with $t\neq 0$, the system $Ax=b$ has no solutions.
You do not need the matrix to have a unique left inverse. In fact, a matrix has a unique left inverse if and only if it is a square matrix and has a two-sided inverse. Matrices that have more rows than columns may have left inverses (if they are of maximum rank) and when they do, they have multiple left inverses.
In summary: if $A$ is square, and it has a left inverse, then it is invertible, it has a unique left inverse, and $Ax=b$ has a unique solution for all $b$. If $A$ has more columns than row, then it never has a left inverse. If $A$ has more rows than columns but is not of maximum rank, then it has no left inverse. And if it has more rows than columns and is of maximum rank, then it has multiple left inverses, and the system $Ax=b$ has either no solutions or exactly one solution, and both situations will occur (since $A$ will not be surjective, so some choices of $b$ will yield a system with no solutions).
