EDIT 23, September 2015
I am updating my answer because my original reply was based on the misunderstanding that the A-series and the B-series are logically equivalent; this misunderstanding being based on an incorrect SEP article on Time which asserts they are "identical".
Your question asks: can a proponent of A-theoretic time avoid the contradiction and associated infinite regress of the A-series by using the B-series. The answer to this question is no. The reason is that a proponent of A-theoretic time rejects B-theoretic time as an invalid representation of time; the two are mutually exclusive. And furthermore, the contradiction is not a feature of B-theoretic time - in the B-series, there is no way to express those situations which give rise to McTaggart’s contradiction of A-theoretic time. So let’s look at why this is the case.
What McTaggart is arguing is that time is an illusion. He does this by arguing that if time is real, then it must be representable by one of two series, those which he calls the A-series and the B-series. Either time flows or it does not. There are not alternatives. Dynamic, moving time is modeled by the A-series. Static, unmoving time is modeled by the B-series.
A-theoretic time corresponds to our intuitive notion of time as flowing from the future, through the present, and into the past. The contradiction and associated infinite regress of A-theoretic time is described in my original reply. Briefly; no event can be both future and present, past and present, or past and future. Yet time understood by the A-series implies that as time flows and every event will possess all three incompatible properties - future, present, and past. Attempts to resolve this contradiction by using tenses leads to an infinite regress which never successfully eliminates the contradiction.
B-theoretic time describes time a unmoving. The temporal properties of events in time can be regarded as fixed and unmoving. The temporal relationships of “before” and “after” are fixed eternally. When we think of one moment in time as being before another, this relationship does not change. For example, the relationship expressed by “in the year 2015, the year 2100 is in the future” is true now just as it is true in the year 9999. McTaggart argues that the B-series in not sufficient for an understanding of what time is. According to McTaggart, we know more about time than just the information represented in the B-series. In fact, the concept of “the present” is not part of the vocabulary of B-theoretic time. This is because, McTaggart argues, that a being capable of knowing (and only knowing) all B-theoretic facts, would know all of the B-series relationships between moments in time, but would not be able to tell you today’s date since he would not be able to identify the present moment. The “present” is only part of the A-series vocabulary.
To recap, McTaggart would consider your question to be ill-posed. Eliminating the contradiction of the A-series by using the B-series is not an option.
ORIGINAL ANSWER
According to Graham Priest, McTaggart argued that the notion of past and future and inherently contradictory, and then concluded that there can be no time. When McTaggart attempts to eliminate this contradiction, he gives rise to an infinite regress. For the sake of completeness, we will look at the contraction, the infinite regress, and the resolution. If you are already familiar with McTaggart's argument, then please skip to the second section, below. (That probably means you, OP.)
First, let's look at why McTaggart claims that the notions of past and future are contradictory.
Let us write P for "it was the case that", and write F for "it will be the case that" - i.e., P for past, F for future. So, for example, if e is some instantaneous event, then we write P(e) to express that "it was the case that e", and similarly for F. Note that e is an instantaneous event - for example, e could denote the moment I click on the "Post Your Answer" button when posting this answer. Now, let h denote the statement "e is occuring". Then we have :
¬( P(h) ∧ F(h) ).
But because time flows, before an event happens it has the property F(h), and after it happens we have P(h). Thus, according to McTaggart we have :
( P(h) ∧ F(h) ).
This is McTaggart's contradiction - ( P(h) ∧ F(h) ) ∧ ¬( P(h) ∧ F(h) ).
As McTaggart notes, this argument is not very convincing since an event cannot occur in both the past and future at the same time. It started off as future; became present; and then was past. But what are we saying here? We are compounding tenses. We are saying P(F(h)) - i.e., it was the case that the event will be a future event - and, F(P(h)) - i.e., it will be the case that the event was a past event. McTaggart then argues that this gives rise to the same contradiction that is outlined above. Again, we must attempt to eliminate this contradiction by compounding our tenses, but this attempt gives rise to the same contradiction. This is McTaggart's infinite regress - there is no escaping the contradiction.
We now resolve this infinite regress.
We begin by noting that every situation s(0) comes together with a set of other situations - situations which are either before or after s(0). Assuming that time is one dimensional, we can represent these situations thus:
... s(-2), s(-1), s(0), s(1), s(2), ...
where those situations left of s(0) are before and those to the right are after.
Here, we have P(h) is true in any given situation s only if it is true in some situation to the left of s. Similarly, F(h) is true in a given situation s only if it is true in some situation to the right of s.
Recall that McTaggart's argument was that, given that h has every possible tense, it is never possible to avoid contradiction. Resolving contradictions in one level of complexity for compound tenses only creates the contradictions in the next level of compound tenses.
So how do we resolve this. Suppose that our instantaneous event h is true only at s(0). Then any statement with compound tenses on h is true somewhere. For example, consider F(P(P(F(h)))). This is true at s(-2).
... s(-3) s(-2) s(-1) s(0) s(1) s(2) s(3) ...
h
F(h)
PF(h)
PPF(h)
FPPF(h)
We can do the same for every compound tense composed of P and F in a consistent manner. Thus, McTaggart's argument fails.
EDIT (31 Aug 15)
Re-reading my answer, I note that I have not made it entirely clear how this eliminates the contradiction in the A and B series. According to SEP, the two series are in fact identical - see my comment below.
In the example given, we apply our zig-zag pattern across the various tenses to arrive at a situation where F(P(P(F(h)))) is true - namely s(-2). So it is true at s(-2) and at each situation to the right of s(-2), while it is false at each situation to the left of s(-2). Thus there is not situation where it is both true and false, and the contradiction is eliminated.
The same reasoning applies to any (compound) tense, where, once we have applied the zig-zag patter across tenses to arrive at a situation where the statement is true, if is then true/false to the left/right according to whether the outermost tense is P/F.
