Explicitly generating 1 in an ideal without prime support The Question
Let $R$ be a unital commutative ring, and let $a,b_1,b_2\in R$.  The following is a basic commutative algebra exercise.
Lemma. If $Ra+Rb_1=R$ and $Ra+Rb_2=R$, then $Ra+Rb_1b_2=R$.
Proof. Let $P$ be any prime ideal containing $Ra+Rb_1b_2$.  Since $b_1b_2\in P$, then $b_1\in P$ or $b_2\in P$.  In either case, $P=R$.  Therefore, no prime ideal contains $Ra+Rb_1b_2$, so it is all of $R$.
Its a slick proof, but its also very nonconstructive.  My question is; given $f_1,f_2,g_1,g_2\in R$ such that
$$ f_1a+g_1b_1=1= f_2a+g_2b_2$$
can you construct $f,g\in R$ such that
$$ fa+gb_1b_2=1?$$
I'm willing to be fairly lax in my standards for a 'construction', in that it doesn't have to be a closed formula.  I just don't want it to use an embedding in a hypothetical prime ideal.
My Motivation
My practical interest in this comes from a non-commutative analog of this problem.  I am considering non-commutative $R$ and quasi-commuting elements $a,b_1,b_2$.  That is, $ab_1=\lambda b_1a$ for some unit $\lambda$ (and likewise for other pairs).
I would like to deduce that
$$Ra+Rb_1=R \text{ and } Ra+Rb_2=R \text{ implies } Ra+Rb_1b_2=R$$
Quasi-commuting elements are close enough to commutative that many constructions still work.  However, one tool which does not generalize is primary decomposition.  Therefore, I would like a more explicit commutative proof, in the hopes that it will work in the quasi-commuting case also.
 A: Here is an example to show it can be done. Consider the q- polynomial ring $R=k_q[x,y]$, the free algebra on two variables over a field $k$ with defining ideal $qxy=yx$ where $q \in k^{\ast}$.  (Usually I would be worried about whether $q$ is a root of unity or not but my example doesn't depend on this).
We have $R(1+xy)+Rx=R$ since $1 \cdot (1+xy)+(-q^{-1}y)\cdot x=1$. Similarly (and rather trivially) we have $R(1+xy)+Rxy=R$. In your notation I'm taking $a=(1+xy), b_1=x,b_2=xy$
The left ideal that you are interested in is $R(1+xy)+Rx^2y$, where the order I multiply $b_1$ and $b_2$ doesn't matter as I can cancel the $q$-powers if I want. I would like $f(x,y)$ and $g(x,y)$ such that 
$$1=f(x,y)(1+xy)+g(x,y)x^2y$$
I can take $f(x,y)=1-xy$ and $g(x,y)=q^{-1 } y$ and this will solve the problem. In this ring I think it may always be possible (provided that the elements you start with satisfy the hypotheses, which is quite restrictive), due to the fact you can write things nicely in a standard form as I have done above. Also the graded structure may may play a part in this positive example.
Good question by the way, I had a fun start to the day thinking about it. 
EDIT: I should point out in relation to darij's comment above that in this ring everything "nearly"quasi-commutes, and for homogeneous elements this is true; it is only when we consider non-homogeneous elements that we will get different powers of $q$ appearing to stop them quasi-commuting. EDIT 2: In fact I just noticed the second comment about only needing 3 quasi-commuting pairs, which may be what my example shows explicitly.
