<B>Edit, 12/2016.</B>  The original post was long ago, but I had a reason to revisit this some months back when I was asked a similar question.  I am including below the positive answer to this question that I found in a special case.  There is a little setup required.  

<B>Flatness over $\mathbb{A}^r$.</B>  First there is a flatness lemma.  Let $k$ be a commutative, unital ring (for instance, $k=\mathbb{Z}$).  Let $A$ be a commutative, unital $k$-algebra that is Noetherian.  Let $a\in A$ be an element.   By the Noetherian hypothesis, the following chain of ideals stabilizes, $$\{0\}=\text{Ann(1)}\subset \text{Ann}(a^1)\subset \text{Ann}(a^2) \subset \dots \subset \text{Ann}(a^n) \subset \text{Ann}(a^{n+1}) \subset \dots $$  There exists a least nonnegative integer $n$ such that $\text{Ann}(a^n) = \text{Ann}(a^{n+1})$.   Of course $n$ equals $0$ if and only if $a$ is a nonzerodivisor in $A$, and in that case $n$ is a zerodivisor in $A$ (since $n$ equals $0$).  The following lemma proves that Moret-Bailly's examples above are definitely a positive characteristic phenomenon.

<B>Lemma.</B> If the integer $n$ is a nonzerodivisor in $A$, then for every $k$-derivation $\delta:A\to A$, $\delta(a)$ is a zerodivisor.  

<B>Proof.</B> This is basically the same proof as in the existence of primary decompositions for Noetherian modules.  Let $b$ be an element in $\text{Ann}(a^n)$ that is not in $\text{Ann}(a^{n-1})$.  Since $a^nb$ equals $0$ in $A$, $-na^{n-1}b\delta(a)$ equals $a^n\delta(b)$.  Since $a^nb$ equals $0$, $$a^{n+1}\delta(b) = -n(a^nb)\delta(a) = 0.$$  Since $\delta(b)$ is in $\text{Ann}(a^{n+1}),$ already $\delta(b)$ is in $\text{Ann}(a^n)$.  Thus, $a^n\delta(b)$ equals $0$, i.e., $(-na^{n-1}b)\delta(a)$ equals $0$.  By hypothesis, $a^{n-1}b$ is nonzero.  Since also $n$ is a nonzerodivisor in $A$, $-na^{n-1}b$ is nonzero.  Thus, $\delta(a)$ is a zerodivisor in $A$. <B>QED</B> 

Let $r>0$ be an integer (the letter "d" is needed for differentials).  Denote by $S_r$ the polynomial $k$-algebra $k[s_1,\dots,s_r]$.  
Let $(a_1,\dots,a_r)\in A^r$ be a collection of elements, and let $f:S_r\to A$ be the unique $k$-algebra homomorphism with $f(s_i)=a_i$ for every $i=1,\dots,r$.  There is an associated $A$-module homomorphism $f^*:\Omega_{S_r/k}\otimes_{S_r} A \to \Omega_{A/k}$.  

<B>Proposition.</B> If $A$ contains $\mathbb{Q}$, and if there exists an $A$-linear retraction $\rho:\Omega_{A/k}\to \Omega_{S_r/k}\otimes_{S_r} A$ of $f^*$, then $(a_1,\dots,a_r)$ are a regular sequence.  In that case, assuming also that $k$ is Noetherian and that $A$ is $k$-flat, then $f$ is a flat ring homomorphism.

<B>Proof.</B>  This is proved by induction on $r$.  First of all, composing the retraction $\rho$ with projection onto the summand $Ads_r$, there exists a $k$-derivation $\delta:A\to A$ with $\delta(a_r)$ equal to $1$.  Thus, by the previous lemma, $a_r$ is a nonzerodivisor.  When $r$ equals $1$, this proves that the sequence is a regular sequence.  Assuming further that $k$ is Noetherian and $A$ is $k$-flat, by applying the Local Flatness Theorem, Theorem 22.2, p. 174 of Matsumura's "Commutative ring theory", also $k[t_1]\to A$ is flat.  Thus, by way of induction, assume that $r>1$, and assume that the result is proved for $r-1$.    

Form the quotient ring $\overline{A}=A/\langle a_r \rangle$ and the image sequence $(\overline{a}_1,\dots,\overline{a}_{r-1})\in \overline{A}^{r-1}$.  Form the projection onto the first $r-1$ summands, $$\Omega_{S_r/k}\otimes_{S_r} A \to \Omega_{S_{r-1}/k}\otimes_{S_{r-1}} A \to \Omega_{S_{r-1}/k}\otimes_{S_{r-1}}\overline{A}.$$  Using the fundamental exact sequence,
$$
\overline{A}\xrightarrow{d(a_r)} \Omega_{A/k}\otimes_A \overline{A} \to \Omega_{\overline{A}/k} \to 0,
$$
the projection above factors through a retraction, $$\overline{\rho}:\Omega_{\overline{A}/k} \to \Omega_{S_{r-1}/k}\otimes_{S_{r-1}}\overline{A}.$$  Thus, by the induction hypothesis, $(\overline{a}_1,\dots,\overline{a}_{r-1})$ is a regular sequence in $\overline{A}$.  Thus, by definition, also $(a_1,\dots,a_r)$ is a regular sequence in $A^r$.  Assuming further that $k$ is Noetherian and $A$ is flat, then $A/\langle a_r\rangle$ is flat over $S_r/\langle s_r \rangle = S_{r-1}$.  Thus, by the Local Flatness Criterion once more, $A$ is flat over $S_r$.  So the propostion is proved by induction on $r$. <B>QED</B>

For any rank $r$, free summand of $\Omega_{A/k}$, since $\Omega_{A/k}$ is generated by symbols $da$ for $a\in A$, at least up to passing to a finite Zariski open cover of $\text{Spec}(A)$, we can assume that there exist elements $a_1,\dots,a_r$ such that the retractions of $d(a_1),\dots,d(a_r)$ form a basis for the summand.  Let $f$ be the associated ring homomorphism.  By the proposition, $f$ is flat.

That is enough to conclude the local product structure, at least when the fibers of $f$ have isolated singularities (as in my proposed counterexample).

<B>Hypothesis.</B> Assume that $k$ is a field containing $\mathbb{Q}$, assume that $f$ is locally finitely presented, and assume that the closed fibers of $f$ have only isolated singularities.



Here is the e-mail I sent to a colleague about this situation.

Dear ***,

As soon as you left, I wrote out the computation for 4 concurrent lines, and I saw that you are (of course) correct: the extension class is zero if and only if the cross-ratio of the four lines is constant.  Then I had to teach my class.  I just finished, and I remembered another reference. That
reference quickly leads to a proof of the formal local splitting whenever
you have vanishing of the Ext class, at least in characteristic 0, at least for germs of isolated singularities.  The reference is in the following book of Michael Artin.

Lectures on Deformations of Singularities <br>
Volume 54 of mathematics: Tata Institute of Fundamental Research lectures <br>
Michael Artin, C. S. Seshadri, Allen Tannenbaum <br>
Tata Institute of Fundamental Research, 1976 <br>
http://www.math.tifr.res.in/~publ/ln/tifr54.pdf

In Theorem 8.1, Artin establishes that there is a versal deformation space
$\text{Spf}(R)$ for a germ of an isolated singularity $(X,0)$ with $X$ locally finitely
presented over a field.  Earlier, on pp. 26, 27, in Theorem 6.2 and
Proposition 6.1, Artin proves that for a flat, finitely presented morphism,

$$f : X \to \text{Spec}(A),$$

with a section,

$$s: \text{Spec}(A) \to X,$$

that is a family of isolated singularities of $f$, for the induced morphism to
the versal deformation space,

$$g : \text{Spec}(A)^{\wedge} \to \text{Spf}(R), $$

the induced derivative is given by the map from the global Ext^1 you were
describing to a natural target space T^1 introduced by Schlessinger.  In
particular, if your Ext class is zero, the derivative is zero.

Of course a nonzero morphism can have a zero derivative at one point.
However, by generic smoothness / Sard's Theorem, in characteristic $0$ a
$k$-morphism from a smooth $k$-scheme to an arbitrary (locally finite type)
$k$-scheme is constant if the derivative map is everywhere zero.  Perhaps you
can find a counterexample in positive characteristic $p$ by considering the
family of concurrent lines, $\text{Zero}(xy(y-x)(y-t^p x))$ in $\mathbb{A}^2 x \mathbb{A}^1$ with
coordinates $((x,y),t)$, where $\mathbb{A}^1$ is the base of the deformation.

For non-isolated singularities, I am not sure what is the best way to
proceed.  Often there is no versal deformation space (at least not as the
formal completion of a finite type scheme at a point), but perhaps the
proofs of Theorem 6.2 and Proposition 6.1 still apply.  You may be able to
find something in the book on deformations of singularities by Greuel,
Lossen, and Shustin.

Jason Starr, April 2016.

<B>Original answer.</B> <I>I am adding below my original (wrong) answer, as well as the edit I made right after posting the answer.</I>
I believe this is false.  There are equisingular families that are not &eacute;tale locally, nor even formally locally, trivial.   Let $(z,w)$ be coordinates in $\mathbb{A}^2$.  Let $t$ be a coordinate on some dense Zariski open subset $V$ of $\mathbb{A}^1$.  Consider the zero scheme $X$ of $zw(w-z)(w-tz)$ in $\mathbb{A}^2\times V$.  I believe that $\Omega_X$ will have a direct sum decomposition as above for $V$ a suitable open subset.  However, I very much doubt there is any &eacute;tale decomposition of $X$.  If there were, then the fiber of the projection to $Y$ would correspond to the zero locus of $(z,w)$.  However, the formal completion of $X$ along this zero scheme is not a product, not even after making a further &eacute;tale base change of $V$.  

<B>Edit.</B> I just did the computation for my polynomial above.  There is not a direct sum decomposition of $\Omega_X$.  I was wrongly assuming that the existence of a direct sum decomposition would be preserved under small deformations.  Thus, deforming from a trivial family to an equisingular family would give a counterexample.  However, because the Ext group is nonzero, existence of a direct sum decomposition is not preserved under small deformations.