Let $A$ be a non-unital $C^\ast$ algebra and let $M(A)$ denote the multiplier algebra and let $\widetilde{A}$ denote the unitisation of $A$.
Consider the map $\varphi : \widetilde{A}\to M(A)$ defined by $(a,\lambda) \mapsto (L_{a+\lambda}, R_{a+\lambda})$. I want to show that this map is injective but can't quite manage somehow. Let $h: \widetilde{A} \to A \oplus \mathbb C$ be the map $(a,\lambda)\mapsto a +\lambda$. Since $A$ is non-unital it is clear that $A \cap \lambda \mathbb C = \varnothing$ so this map is injective. Next consider the map $g: A \oplus \mathbb C \to M(A)$ defined by $a + \lambda \mapsto (L_{a+\lambda}, R_{a+\lambda})$. I want to show $g$ is injective since then $\varphi = g \circ h$ is, too.
So let $(a,\lambda), (b, \eta) \in A \oplus \mathbb C$ be such that $g(a, \lambda) = g(b, \eta)$. Then for all $c \in A$: $$ L_{a+\lambda}c = (a+\lambda)c = (b+\eta)c = L_{b+\eta}c$$
and therefore $(a+\lambda-b-\eta)c = 0$. Now if $A$ was unital I could use $c=1$ and be done. But $A$ is non-unital, so what do I do?
Let $A$ be a non-unital $C^\ast$ algebra and let $\widetilde{A}$ denote its unitisation. Let $M(A)$ denote the multiplier algebra. Define a map $\varphi : \widetilde{A}: M(A)$ as $(a,\lambda) \mapsto (L_{a+\lambda},R_{a+\lambda})$. Then $\varphi$ is injective.
Note that $\varphi$ is linear. Hence it is enough to show that $\varphi(a, \lambda)=0$ implies that $(a,\lambda) = (0,0)$.
If $\lambda = 0$ and $\varphi (a,0) = L_a$ is the zero map then $L_a b = 0$ for all $b \in A$. In particular, $L_a(a^\ast) = aa^\ast = 0$. Then
$$ 0= \|aa^\ast\| = \|a^\ast a\| = \|a\|^2 $$ and hence $a=0$.
If $\lambda \neq 0$ then since $A$ is not unital there does not exist $a\in A$ such that $a = -\lambda \cdot 1$. Hence $L_{a+\lambda}\neq 0$ for all $a \in A$.
Hence $\varphi (a, \lambda) = 0$ implies that $(a,\lambda ) = (0,0)$ and therefore $\varphi$ is injective.