According to every definition in the literature, (1) a homomorphism of DG-modules must be a chain map [Stacks Project]. This is perfectly reasonable at first glance, especially if you want to use the forgetful functor to chain complexes.
It also seems reasonable, or so I thought, that (2) a DG-algebra $A$ is a right (and left) DG-module over itself. In particular, for every $a\in A$ it seems reasonable that $\mu_{a}: A\to A$, $\mu_{a}(b) = a * b$ should be a right DG-module homomophism.
Claim: Properties (1) and (2) are incompatible.
Proof by example. Consider a DG-algebra $A$ which is an object in the image in the category of DG-algebras of the monoidal Dold-Kan correspondence [nLab]. Here is the most concrete nontrivial example:
Let $A_{0}, A_{1}$ be subrings of a commutative ring $A_{01}$. Let $A^{0} = A_{0} \oplus A_{1}$, and $A^{1} = A_{01}$, and $A = A^{0} \oplus A^{1}$. Define $d(a_{0},a_{1},a_{01}) = (0, 0, a_{1} - a_{0})$, and the DG-algebra multiplication to be:
$$ a * b = (a_{0}, a_{1}, a_{01}) * (b_{0}, b_{1}, b_{01}) = (a_{0} b_{0}, a_{1} b_{1}, a_{0} b_{01} + a_{01} b_{1}) $$
Where the products $a_{*}b_{*}$ happen inside $A_{01}$.
Claim: $A$ is a DG-algebra, but $\mu_{a}$ is not a chain map.
Proof: For simplicity choose $|a| = 0$, so $a = (a_{0}, a_{1}, 0)$. Then
$$ d(\mu_{a}(b)) = d(a * b) = d(a_{0} b_{0}, a_{1} b_{1}, a_{0} b_{01}) = (0, 0, a_{1} b_{1} - a_{0} b_{0}) $$ but $$ \mu_{a}(d(b)) = \mu_{a}(0, 0, b_{1} - b_{0}) = (0, 0, a_{0}(b_{1} - b_{0}) ) $$
This is rarely a chain map. The failure of it to be a chain map is measured by $da$:
$$ d(\mu_{a}(b)) - \mu_{a}(d(b)) = (0, 0, (a_{1} - a_{0}) b_{1}) $$
Am I crazy here? Because of the application I need them for, I would much rather have (2) be true and dispose of property (1) from the definition of a DG-module homomorphism.
This is incorrect. The correct claim is that if $a$ is a $0$-cycle then $\mu_a$ is a dg module homomorphism. Explicitly, we have
$$d(ab) = d(a) b + (-1)^{|a|} a \, d(b)$$
and so if $|a| = da = 0$ then $d(ab) = a \, d(b)$, making multiplication by $a$ a chain map.
More generally, multiplication defines a morphism
$$A \to [A, A]$$
of chain complexes, where $[A, A]$ is the internal hom of chain complexes. Homomorphisms are given by $0$-cycles in the internal hom, and more generally morphisms of degree $n$ are given by $n$-cycles.