Let $A$ and $B$ be two $\ast$-algebras and let $\varphi:A\to B$ be a $\ast$-algebra morphism.
I am interested in all the ways that $\varphi$ could fail to be unital (that is, if $A$ has a unit $1_A$, how could it be that $\varphi(1_A)\neq1_B$?).
Using the same proof as in group theory, it seems like if both algebras contain only invertible elements, then $\varphi$ is always unital: $$ \varphi(1_A)=\varphi(a\cdot a^{-1})=\varphi(a)\cdot\varphi(a^{-1})=\varphi(a)\cdot\varphi(a)^{-1}=1_B$$where in the first equality we have used the fact that $A$ contains an invertible element $a$ and in the third equality the fact that a group morphism preserves inverses.
So my question is:
How else could $\varphi$ fail to be unital except for these cases:
1) $A$ does not have a unit.
2) $A$ does not have any invertible elements.
Let $A=\left\{\begin{pmatrix} a&0\\0&b \end{pmatrix}\mid a,b\in \mathbb{C}\right\}$ and $B=\left\{\begin{pmatrix}a&0&0\\0&b&0\\0&0&c\end{pmatrix}\mid a,b,c\in \mathbb{C}\right\}.$ Then $A$ and $B$ are two unital $*$-algebras in the usual way. Consider the map
$$\phi:A\rightarrow B:\begin{pmatrix} a&0\\0&b \end{pmatrix}\mapsto \begin{pmatrix} a&0&0\\0&b&0\\0&0&0 \end{pmatrix}.$$ Then $\phi$ is a $*$-algebra morphism that is not unital.
Edit: Notice however that $\phi(1_A)$ is the unit of $\text{Im}(\phi)$.