Yes, this can be done without making the $C^*$-algebra concrete in $B(H)$. All we need is that $z=0$ iff $z^*z=0$, which follows from the $C^*$-identity $\|z^*z\|=\|z\|^2$.
Lemma Let $a\in A$. The following assertions are equivalent.
1 - $p=a^*a$ is idempotent, i.e. $p^2=p$.
2 - $aa^*a=a$
3 - $a^*aa^*=a^*$
4 - $q=aa^*$ is idempotent.
$1\Rightarrow 2:$ Here is the main trick. Set $z:=aa^*a-a$. Note that $z^*z=p^3-p^2-p^2+p=0$, whence $z=0$.
$2\Rightarrow 3:$ Take the adjoint.
$3\Rightarrow 4:$ Left-multiply $3$ by $a$.
$4\Rightarrow 1:$ Applying $1\Rightarrow 2$ to the element $a^*$ instead of $a$, we get $a^*aa^*=a^*$. Right-multiplication by $a$ yields $1$. $\Box$
By definition, an element $a$ is a partial isometry in an abstract $C^*$-algebra if $a^*a$ is a projection. Of course, the latter is always self-adjoint. So $a$ is a partial isometry iff $a^*a$ is idempotent. It follows in particular that $a$ is a partial isometry iff $a^*$ is a partial isometry.
Remark Two projections $p,q$ as above are called Murray-von Neumann equivalent. The lemma above helps prove that the latter is an equivalence relation (namely transitivity, which is the slightly tricky part).