Deformation of relations
Answer to question 2 is the following: a deformation of an algebra $A_0$ parametrized by a pointed affine scheme $*\to X=Spec(B\to k)$ is the data of a $B$-algebra $A$ such that $A_0\cong A\otimes_B k$.
Observe that the kind of deformations your are looking at in question 1 are those that are called "flat".
Deforming structure constants of matrix algebras
In order to find the answer to question 1 (which is yes) I would suggest you to use Artin–Wedderburn theorem.
EDIT 1: more precisely, over $\mathbb{C}$ any finite-dimensional (semi-)simple algebra is isomorphic to (a Cartesian product of) matrix algebras.
EDIT 2: Anyway, it seems that there is a more general phenomenon (see the fondational paper of Gerstenhaber, On the deformation of rings and algebras, Ann. of Math. 79 (1), (1964),
59–104):
$$
\textrm{absolute rigidity}\Rightarrow \textrm{analytic rigidity}\Rightarrow \textrm{geometric rigidity}
$$
where absolute rigidity means that $HH^2(A)=0$, analytic rigidity means that there are no non-trivial formal deformations of $A$, and geometric rigidity means that the $GL(V)$-orbit of the point defined by $A=(V,c_{ij}^k)$ in the variety of structure constants on $V$ is open (this tells you in particular that sufficiently small deformations are trivial). You can conclude by using that
$$
\textrm{semi-simplicity}\Rightarrow\textrm{absolute rigidity}
$$
Azumaya algebras
Finally, if I understand well your question 3, your guess seems correct: the right notion to look at is precisely the one of Azumaya algebra (if you are interested by the relation to Cherednick algebras mentionned in the answer of Daniel Larsson, I would suggest you to take a look at the last two or three chapters of this book by Pavel Etingof).
Final remark
Let me also mention that there is a lot of work about deformation theory of quadratic algebras by sub-quadratic ones (see references given in this MO answer - EDIT: sorry, you already know this since you're the one who actually asked the question!)
To add a bit to what Damien says, addressing your question on how to generalise the gauge approach (which is equivalent to the approach outlined by Damien, as proved by several people):
You can view gauge symmetries in DGLAs via solving the differential equation
$$
\frac{d\alpha}{dt}=-\partial\xi-[\alpha,\xi],
$$
where $\xi$ is the given element of degree $0$. This generalises to homotopy Lie algebras as follows: consider the differential equation
$$
\frac{d\alpha}{dt}=-\partial\xi-[\alpha,\xi]-\frac12[\alpha,\alpha,\xi]-\ldots-\frac{1}{p!}[\underbrace{\alpha,\ldots,\alpha,}_{p \text{ times}}\xi]_{p+1}-\ldots,
$$
where the right hand side is simply the negative of $[\xi]_1^\alpha$, the first structure map of the twisted Lie-infinity structure
$$
[x_1,\ldots,x_k]_k^\alpha:=\sum_{p\ge0}\frac{1}{p!}[\underbrace{\alpha,\ldots,\alpha,}_{p \text{ times}}x_1,\ldots,x_k]_{k+p}.
$$
From that it is almost obvious that moving along the integral curves of this equation preserves the property of being Maurer--Cartan, since the Maurer--Cartan condition for $\alpha+\beta$, where $\alpha$ is a Maurer--Cartan element, and $\beta$ is infinitesimal becomes
$$
\partial\beta+[\alpha,\beta]+\frac12[\alpha,\alpha,\beta]+\ldots+\frac{1}{p!}[\underbrace{\alpha,\ldots,\alpha,}_{p \text{ times}}\beta]_{p+1}+\ldots,
$$
that is $[\beta]_1^\alpha=0$, and so $\beta=[\xi]_1^\alpha$ satisfies that, $[\cdot]_1^\alpha$ being a differential of the twisted structure. This circle of ideas is explained in many places, one important reference is ``Lie theory for nilpotent $L\_\infty$-algebras'' by Ezra Getzler (Ann. of Math. (2) 170 (2009), no. 1, 271--301.).
Best Answer
A good way of rewriting the gauge action is $e^{f}(\mu+\mu')e^{-f}=\mu+e^{f}*\mu'$, which makes the equivalence manifest. There are several references for the computation above. The first coming to my mind is Marco Manetti's Lectures on deformations of complex manifolds where, if I'm not wrong, it's spelled out in detail.