Banach algebra embedding into the algebra of bounded operators on it 
Suppose $(A, \|\cdot\|, *)$ is a non-unital Banach algebra, where $\|\cdot\|$ is the norm. Then consider the injective Banach algebra homomorphism $\iota: A \hookrightarrow \mathcal{B}(A)$, defined by $\iota: a \mapsto (x\mapsto a*x)$, where we denote $\hat{a}:=\iota(a): x\mapsto a*x$ as the corresponding left multiplication map from $A$ to itself, for all $x\in A$.

Then, by Banach algebra inequality, we have
$\|\hat{a}\|_{\mathcal{B}(A)}=\sup \frac{\|a*x\|}{\|x\|}\leqslant \|a\|<\infty.$
So embedding $\iota$ is bounded, since we always have $\|\iota\|_{\text{OP}}=\sup \frac{\|\hat{a}\|_{\mathcal{B}(A)}}{\|a\|}\leqslant 1$.

Question 1: Under what conditions is $1_{\mathcal{B}(A)}$, the identity map, in the closure of the image $\iota(A)$ in $\mathcal{B}(A)$?


Question 2: In case that happens, we can always have a net $(\hat{\mu_\lambda})_{\lambda}\subset \iota(A)\subset \mathcal{B}(A)$ such that $\|\hat{\mu_\lambda} - 1_{\mathcal{B}(A)}\|_{\mathcal{B}(A)}\to 0$. Under what conditions, is its preimage $(\mu_\lambda)_\lambda\subset A$ an approximate unit of $A$?


Question 3: With all the above assumptions, can the embedding $\iota$ tell us when algebra $A$ admits a sequential approximate unit?

Some explanation:
In a recent question, I asked about why we can't always have sequential approximate units in general non-unital Banach algebras. It seems, the answers I've got there can't really explain the nature of this phenomenon. But if we think of it in the context of the above embedding, we seem to be able to understand better.
A key issue in QUESTION 2 is, we only have
$$\|\mu_n * x-x\|\geqslant \|\hat{\mu_n} \circ \hat{x} -\hat{x}\|_{\mathcal{B}(A)}\to 0,$$
which cannot guarantee the left-hand side also converge to $0$. So we have to consider all the nets $(\hat{\mu_\lambda})_\lambda\subset \iota(A)\subset \mathcal{B}(A)$ which converge to $1_\mathcal{B}(A)$. Among them, there might be some nets (or maybe not), whose preimage $(\mu_\lambda)_{\lambda}\subset A$ satisfy the definition of an approximate unit.
 A: If $a$ is an element of $A$, then the map $i(a):x\in A\mapsto ax\in A$ is clearly a map of right $A$-modules. A strong limit of maps of right $A$-modules is a map of right $A$-modules, so if the image of $i$ is dense, then all maps in $B(A)$ are right $A$-linear.
Now if $b$ is an element of $A$, then map $j(b):x\in A\mapsto xb\in A$ is bounded, so we see that it is right $A$-linear: this implies that $b$ is central in $A$, and this that the algebra $A$ is commutative.
Since the map $a\in A\mapsto i(a)\in B(A)$ is a map of algebras with dense image, this implies that $B(A)$ is commutative, and therefore $A$ has dimension $1$.
A: The answer is based on
Lemma For a Banach space, if $T\in B(X)$ is not a multiple of $I,$ there exists $S\in B(X),$ such that $TS\neq ST.$
Let $\dim(A)\ge 2.$ Assume  that  $l(A)$ is dense in $B(A).$ Then the operators of the form $r(a)x=xa$ commute with all operators in $B(A),$ as they commute with $l(b)$ for any $b\in A.$ Fix $a\in A$ such that $r(a)$ is not a multiple of the identity operator (if such $a$ does not exists the algebra is one-dimensional). Then, applying the lemma there is $S\in B(A)$ such that $Sr(a)\neq r(a)S,$ a contradiction.
A: Let the Banach Algebra $A$ be finite ($n$) dimensional. In this case the question is does $A$ admit a representation in the form a $n \times n$ matrix over the base field. If $\{v_1,....,v_n\}$ are the basis then we can write left multiplication by an element $a$ as a matrix $[a.v_1,a.v_2,...,a.v_n]$. Hence each element $a \in A$ has a matrix representation: $[a.v_1,a.v_2,...,a.v_n]$. The question is does this matrix uniquely represent an element $a \in A$. If not then $[(a_1-a_2).v_1,(a_1-a_2).v_2,...,(a_1-a_2).v_n] = 0$ implying that $(a_1-a_2).x = 0$, $\forall x \in A$. Since $A$ is non-unital, its not clear that $a_1=a_2$. May be you want to start with this. If $A$ doesnt have zero divisors then $a_1=a_2$ and hence its enough to look at matrix algebras for your case when $A$ doesnt have zero divisors. check under what conditions $a.x = 0$, $\forall x \in A$ implies $a=0$. This is a weaker condition than requiring no zero divisors. Let $a=\sum_i g_i v_i$. Now $a.x = 0$ implies, $a.v_j=\sum_i g_i v_i.v_j = 0$. Hence the matrix $M_k=[v_{ij}]$ with $v_{ij} = v_k.v_j$ is a singular matrix. Under these conditions the matrix representation need not be unique and when $M_k$ for some $k \in [n]$ is non-singular then matrix representation is unique. If matrix representation is possible then your question boils down to analyzing matrix subalgebras over fields. Interesting question.
