Is it generally implicitly assumed that if $B$ is a subalgebra of a unital Banach algebra $A$ then $1 \in B$?
I tried to find a definition of subalgebra but the only definition I found was in Murphy who defines it to be ''itself a normed algebra with the norm got by restriction''
It seems to be that here at least $1 \in B$ is not required. It's also easy to come up with examples: Let $A = \mathbb C^2$ and $B=$ one the axes (unit $=(1,1)\notin B$).
Nevertheless, I have doubts. Are there situations where it is desirable to consider non-unital subalgebras of unital algebras?
First, an answer to your question. No, one typically does not assume that Banach subalgebras are unital. One of the standard examples of a Banach algebra is $C(X)$ where $X$ is compact, and one of the standard examples of a Banach subalgebra of $C(X)$ (an ideal, in fact) is is $C_0(U)$ where $U \subseteq X$ is an open set. But $C_0(U)$ is not unital.
Second, a comment. In Banach algebra theory you do not need to worry about units the way you do in most of ring theory simply because unitalizations behave very nicely. If $A$ is a unital Banach algebra and $B$ is a non-unital Banach subalgebra the unitalization $\tilde{B}$ sits naturally in $A$ (not true for arbitrary rings, even commutative ones).