left invertible elements in a ring with unity. My question is:
Let $R$ be a ring with the unity $e$ and $a \in R$. If $a^{\circ}\triangleq\{x\in R \;| \;ax=0\}=\{0\}$, as the following conter-example given by Matthias Klupsch, we know that it is not left invertible in general. Then what additional conditions on $R$ or on $a$ will make that $a$ is left invertible?
If we have known $a$ is right invertilbe, then what will happen?
${\bf Notes:}$   Let $A$ and $B$ be two sets, and $f : A \to B$ be a mapping, then it is well-known that:
(1) $f$ is injective iff $f$ is left invertible.
(2) $f$ is surjective iff $f$ is right invertible.
Thanks!
 A: This question can also be considered in categorical setting: a ring $R$ itself is a (pre-)additive category with one object, meaning that it has $+,-$ operations (and of course $0_{AB}$ arrow) on the hom-sets (now we have only one hom-set for the one object).
In preadditive category, $ax=ay \implies a(x-y)=0 \overset{\text{cond.}}\implies x-y=0 \implies x=y$, hence your condition means exactly that $a$ is left cancellable.
So, basically, you are asking whether left cancellable elements ('epimorphic' if the composition is read from left to right)  are left invertibles as well. And the answer is: no in general. (Matthias and others wrote nice simple counterexamples in the comments.)
A: If you look at the polynomial ring $ R[X] $ over an integral domain $ R $, for any $ p \in R[X] \setminus R $, we have $ p^{\circ} = \{ 0_{R[X]} \} $. However, $ p $ is not invertible. This serves as a counterexample.
With regards to the latest edit, let $ \mathbb{R}^{\mathbb{N}} $ denote the $ \mathbb{R} $-vector space of all sequences of real numbers. Consider the $ \mathbb{R} $-linear map $ F_{\text{Right}} $ on $ \mathbb{R}^{\mathbb{N}} $ that shifts a sequence to the right by a single position. For example,
$$
{F_{\text{Right}}}(a_{1},a_{2},a_{3},a_{4},\ldots) = (0,a_{1},a_{2},a_{3},\ldots).
$$
Clearly, $ F_{\text{Right}} $ has a left-inverse, which is the $ \mathbb{R} $-linear map $ F_{\text{Left}} $ on $ \mathbb{R}^{\mathbb{N}} $ that shifts a sequence to the left by a single position. However, there is no $ \mathbb{R} $-linear map that acts as a right-inverse of $ F_{\text{Right}} $, because all sequences in the range of $ F_{\text{Right}} $ have a $ 0 $ in the first position. This implies that in the endomorphism ring of $ \mathbb{R}^{\mathbb{N}} $, where ring multiplication corresponds to composition of $ \mathbb{R} $-linear maps, there exist elements that are left-invertible but not right-invertible.
Without much pain, one can just as easily produce a ring $ R $ that has right-invertible-but-not-left-invertible elements.
Conclusion In general, left-invertibility does not imply right-invertibility, and right-invertibility does not imply left-invertibility. However, in square-matrix rings over a field, the two imply each other.

Addendum Given a ring $ R $, if $ a \in R $ satisfies (i) $ a^{\circ} = \{ 0_{R} \} $ and (ii) $ a $ is right-invertible, then $ a $ is automatically left-invertible. Indeed, suppose that $ ab = 1_{R} $ for some $ b \in R $. Then
$$
a(1_{R} - ba) = a - a(ba) = a - (ab)a = a - 1_{R} \cdot a = a - a = 0_{R}.
$$
As $ a^{\circ} = \{ 0_{R} \} $, it follows that $ 1_{R} - ba = 0_{R} $, or equivalently, $ ba = 1_{R} $. Therefore, $ a $ is left-invertible.
The foregoing argument shows why right-invertibles in a square-matrix ring over a field must also be left-invertibles. Let $ A \in {\text{M}_{n}}(\mathbb{F}) $ be right-invertible, i.e., $ AB = \mathbf{I}_{n} $ for some $ B \in {\text{M}_{n}}(\mathbb{F}) $. Viewing $ A $ and $ B $ respectively as linear transformations $ T_{A} $ and $ T_{B} $ on $ \mathbb{F}^{n} $, we have $ T_{A} \circ T_{B} = \text{id}_{\mathbb{F}^{n}} $. Hence, $ T_{B} $ is injective and $ T_{A} $ is surjective. However, the Dimension Theorem from linear algebra tells us that surjective linear operators on finite-dimensional vector spaces are also injective, so
\begin{align}
A^{\circ} &= \{ X \in {\text{M}_{n}}(\mathbb{F}) \,|\, AX = 0_{{\text{M}_{n}}(\mathbb{F})} \} \\
          &= \{ X \in {\text{M}_{n}}(\mathbb{F}) \,|\, T_{A} \circ T_{X} = 0_{\mathcal{L}(\mathbb{F}^{n},\mathbb{F}^{n})} \} \\
          &= \{ X \in {\text{M}_{n}}(\mathbb{F}) \,|\, T_{X} = 0_{\mathcal{L}(\mathbb{F}^{n},\mathbb{F}^{n})} \} \\
          &= \{ 0_{{\text{M}_{n}}(\mathbb{F})} \}.
\end{align}
Therefore, $ BA = \mathbf{I}_{n} $, i.e., $ A $ is left-invertible.
A: (It's not clear to me what you were asking in your question about right-invertibility, but perhaps it was this.)  
One positive result is that if $a^\circ = \{0\}$ and if $a$ is right-invertible, then $a$ is invertible. 
Indeed, if $ab = 1$, then $a(1-ba) = a - (ab)a = a - a = 0$. So $1-ba \in a^{\circ}$, which means that $ba = 1 = ab$.  Hence $a$ is invertible with $a^{-1} = b$.
In particular, in a domain (a nonzero unital ring without zero-divisors), every right invertible element is invertible.
