As Mariano noted, you can tell immediately that they are not isomorphic by considering a number of their algebra invariants.
To know that they are isomorphic is usually more complicated. Though note that, as Ofir noted, all abelian groups of the same order over $\mathbb{C}$ (or any field with a primitive root with order the order of the groups in question) have isomorphic group algebras.
The reason for that is that the group algebras can be written as a sum of modules. A necessary and sufficient condition for $kG_1\cong kG_2$ is that they both admit module decompositions which are equivalent as vector spaces: just match up the dimensions and multiplicities of each factor. As an example, over $\mathbb{C}$ the group algebras of $D_8$—the dihedral group of order 8—and $Q$—the quaternion group of order 8—are isomorphic since they have the same number of 1-dimensional and 2-dimensional irreducible representations, and no higher irreducible representations. The high-level way of stating that is to say that the two representation categories have isomorphic Grothendieck rings.
Of course, knowing the $k$-representation theory of the groups in question may not be such a simple thing, but there's no real way of getting around that problem if you want to establish the isomorphism.
There is one (more restrictive) situation for which the question is pretty well-investigated, and that's the situation of isocategorical groups. Two groups are isocategorical when their $\mathbb{C}$-representation categories are monoidally equivalent (importantly, the symmetry structure of the tensor product will not be preserved if the groups are not isomorphic). This is a stronger statement than having isomorphic Grothendieck rings: $Q$ and $D_8$ are not isocategorical. Indeed, the linked paper shows that any group which is isocategorical to a non-isomorphic group necessarily has an abelian normal subgroup of order a power of 4 that can be equipped with a certain isomorphism (the lack of this isomorphism is what stops $Q$ being isocategorical to $D_8$).
For your particular groups, notice that they cannot be isocategorical: they have no abelian normal subgroups a power of 4. They have abelian subroups of order 4, but they're not normal.
So for your two particular groups you're pretty much forced to compute the representation categories (over $\mathbb{C}$, say). Your groups should be relatively easy to compute this for: dealing with the direct products is trivial and the semidirect products should be routine (induce representations of $Q_i$ up). The only potentially hard part is to figure out the representations of $Q_1$ and $Q_2$, but if you've ever tried constructing a character table for a group then that's a good starting point. Note that their commutator subgroups are easy to find, whence their linear character groups are easy to compute.
If $\mu_*$ is not a monomorphism, then there exists a nonzero element that lies in the kernel. But that nonzero element can be expressed as a finite linear combination of pure tensors (elements of the form $a\otimes g$ for $a\in A$ and $g\in G$). Thus, this witness will also exist if we replace $G$ with the subgroup generated by the $g$ that occur in this witness. Thus, we can reduce to $G$ finitely generated.
Since $G$ is torsion free and finitely generated, it is isomorphic to a direct sum of a finite number of copies of $\mathbb{Z}$. But the tensor product distributes over (finite) direct sums
:
$$A\otimes_{\mathbb{Z}}(G_1\oplus G_2) \cong (A\otimes_{\mathbb{Z}}G_1)\oplus (A\otimes_{\mathbb{Z}}G_2).$$
Thus, we may decompose the map $\mu_*\colon A\otimes_{\mathbb{Z}}G \to B\otimes_{\mathbb{Z}}G$ into a family of maps with domain $A\otimes_{\mathbb{Z}}\mathbb{Z}$ and codomain $B\otimes_{\mathbb{Z}}\mathbb{Z}$ (using thee universal property of the direct sum as both a coproduct and a product for abelian groups). If $\mu_*$ is not injective, then at least one of these maps is not injective. Thus, we are reduced to the case of $G\cong \mathbb{Z}$.
So now we have a monomorphism $\mu\colon A\to B$, and the induced map $\mu_*\colon A\otimes_{\mathbb{Z}}\mathbb{Z}\to B\otimes_{\mathbb{Z}}\mathbb{Z}$. We want to verify that this is a monomorphism.
Lemma. For all abelian groups $A$, $A\otimes_{\mathbb{Z}}\mathbb{Z}\cong A$.
Proof. First note that every element of $A\otimes_{\mathbb{Z}}\mathbb{Z}$ can be written as a pure tensor of the form $a\otimes 1$: indeed, given an element
$$\sum_{i=1}^n m_i(a_i\otimes k_i)$$
we can first use bilinearity to rewrite $a_i\otimes k_i=(k_ia_i)\otimes 1$, and then rewrite $m(k_ia_i\otimes 1) = (m_ik_i)a\otimes 1$. Then we have
$$\sum_{i=1}^n m_i(a_i\otimes k_i) = \sum_{i=1}^n (m_ik_ia_i)\otimes 1 = \left(\sum_{i=1}^n m_ik_ia\right)\otimes 1.$$
Consider the map $f\colon A\otimes_{\mathbb{Z}}\mathbb{Z}\to A$ induced by the bilinear map $A\times \mathbb{Z}\to A$ given by $(a,n)=na$. Assume that $x\in A\otimes_{\mathbb{Z}}\mathbb{Z}\in\ker(f)$. Then we write $x=a\otimes 1$ for some $a\in A$ and we conclude that $0=f(x)=f(a\otimes 1)=a$. But $0\times 1=0_{A\otimes_{\mathbb{Z}}\mathbb{Z}}$, so $f$ is one-to-one. Finally, $f$ is surjective, since $a\in A$ is the image of $a\otimes 1$. $\Box$
Now consider the map $A\to B$ given by
$$A\stackrel{\cong}{\to}A\otimes_{\mathbb{Z}}\mathbb{Z}\stackrel{\mu_*}{\to}B\otimes_{\mathbb{Z}}\mathbb{Z}\stackrel{\cong}{\to}B.$$
The map is
$$a \longmapsto a\otimes 1\longmapsto \mu(a)\otimes 1\longmapsto \mu(a).$$
That is, this map is equal to $\mu$. Since $\mu$ is 1 to 1 and both the first and last maps are isomorphisms, $\mu_*$ is also one-to-one.
Which is what we wanted to prove.
Best Answer
Yes, it's just the group algebra of the pushout of $G_1$ and $G_2$ over $G$ (explicitly, the quotient of $G_1\oplus G_2$ by the image of $(\phi_1,-\phi_2):G\to G_1\oplus G_2$). This follows immediately from the fact that the group algebra functor from abelian groups to commutative $k$-algebras is left adjoint to the multiplicative group functor, and thus preserves colimits, so it turns the pushout of $G_1$ and $G_2$ over $G$ into the pushout of $k[G_1]$ and $k[G_2]$ over $k[G]$ which is just the tensor product.