Specific basis of A-algebra B that is also a free A-module of finite rank. I have a problem that seems (at least to me) harder then I initially thought. 

Let $B$ be an $A$-algebra that is also a free $A$-module of finite rank (if necessary we can assume that $B$ is normal). Can we find an $A$-basis of $B$ that contains $1$?

Since $B$ is free we have a basis, say $\{b_1,\ldots,b_n\}$. We can write $1=\sum a_ib_i$ for some $a_i \in A$. If we could proof that one of the $a_i$ is invertible we would be done, since we could just exchange $1$ for the corresponding $b_i$. But I couldn't do it.
Some help or a counterexample would be appreciated.

Edit: I am most interested in the case where $A=\mathbb{Z}$ and $B$ is a integral extension of $A$. But I'm also keen on seeing this in a more general setting.
 A: Slup has given you the answer where it is mostly used, though it is not in general true that Trace maps $B$ to $A$, unless you assume something more (typically, one assumes that $A$ is integrally closed).
One has standard counterexamples for general cases. For example take a ring $A$ which has a non-free projective module $P$ such that $A\oplus P$ is free. (Standard example is take $A=\mathbb{R}[x,y,z]/(x^2+y^2+z^2-1)$, the coordinate ring of the real sphere and $P$ to be its tangent bundle, which is a rank 2 non-free projective module over $A$ and $A\oplus P$ is free.) Now, $B=A\oplus P$ has a natural ring structure, $(a,p)(b,q)=(ab, aq+bp)$ with $j:A\to B$ being $j(a)=(a,0)$. You can check easily now that $j(1)=(1,0)$ can not be taken as part of a basis for $B$.
A: This does not cover the full range of your question but at least gives an affirmative answer to an interesting special case.
I am denoting the structural morphism by $j:A\rightarrow B$.
Suppose first that $A$ is an integral domain. If $B$ is a free $A$-module, then for every nonzero element $b\in B$ we have $ab=0$ for some $a\in A$ if and only if $a=0$. So $j:A\rightarrow B$ is always injective.
Suppose now that $A$ and $B$ are integral normal domains and assume that the induced extension of their fraction fields  $K\subseteq L$ is separable. Consider the trace map:
$$\mathrm{Tr}_{L/K}:L\rightarrow K$$
which restricts to a morphism of $A$-modules:
$$\mathrm{Tr}_{B/A}:B\rightarrow A$$
Since $K\subseteq L$ is separable, you can check that:
$$\mathrm{Tr}_{B/A}\cdot j=1_A$$
Hence we derive that:
$$B=M\oplus j(A)$$
Edit: This is not the end of the argument as it was pointed out in the comments below. One should impose some extra assumptions on $A$ in order to know that $M$ is free. This holds for some special classes of rings like:
PID, semilocal rings and polynomial rings over fields.
For PID this is standard, for semilocal rings it holds due to the fact that every finitely generated projective module of constant rank is free and finally for polynomials over a field every finitely generated projective is free. The last fact mentioned is highly nontrivial:
https://en.wikipedia.org/wiki/Quillen-Suslin_theorem
