Let $A$ be a non-unital $C^{\star}$-algebra and let $B(A)$ be the Banach algebra of bounded linear operators from $A$ to itself. For $a\in A$ we consider the left-multiplier $L_{a}\colon A\to A$ defined by $L_{a}(b):=ab$. I am trying to prove that $$\phi\colon A\oplus\mathbb{C}\to B(A),\qquad\phi(a,\lambda):=L_{a}+\lambda I$$ is an injective (linear) map.
If $L_{a}+\lambda I=0$, then $ab+\lambda b=0$ for all $b\in A$. So I was looking for $b$'s that yield $a=0$ and $\lambda=0$. If I can first prove that $\lambda=0$, then it is easy to see that $a=0$ by considering $b:=a^{*}$ ($C^{\star}$-identity).
Thus I need to find a $b$ that guarantees $\lambda=0$. I can't figure out this seemingly easy looking problem. Any help would be greatly appreciated!
EDIT: It seems that I haven't yet used the fact that $A$ is non-unital.
Suppose $\lambda\ne 0$. Then $ab+\lambda b=0$ implies $b=(-\frac{a}{\lambda})b$ for all $b\in A$. So if we let $e=-\frac{a}{\lambda}$ then $e$ is a left identity of $A$. By taking involution on the equality $b^*=eb^*$ we also obtain $b=be^*$ for all $b\in A$, which means $e^*$ is a right identity of $A$. Now, using the standard trick:
$e^*=ee^*=e$
And so $e$ is a two sided identity of $A$, a contradiction.