Does isomorphism mean "same algebraic structure"? Let $(M_1,+,\times)$ be an algebraic structure, lets say, for example, a ring. If we have another structure $(M_2,+,\times)$ isomorphic to the first one does that mean that $(M_2,+,\times)$ is also a ring ?
Edit : To give more context, in 2 different exercises I had to prove that the following are rings :
1) $(\mathbb{C},+,•)$ where "$+$" is simple addition and $z_1•z_2 = z_1z_2 + \operatorname{Im}(z_1) \operatorname{Im}(z_2)$, where $\operatorname{Im}(z) = $ the imaginary component of z.
2) $(\mathbb{R} \times \mathbb{R},+,•)$ where $(a,b)+(x,y) = (a+x,b+y)$ and  $(a,b)•(x,y) = (ax,ay+bx)$ 
In fact, these are the same with different notations:  $x+yi = (x,y)$
 A: Sometimes.
It depends on what aspects of the two algebras are isomorphic; for example, $(M_1, +_1)$ could be isomorphic to $(M_2, +_2)$ as groups, but not necessarily as, say, rings $(M_1, +_1, \times_1)$ and $(M_2, +_2, \times_2)$. If they're isomorphic as the same structure, including the ring axioms, then, yes, if one is a ring, so is the other.
A: I think what you are looking for is the idea of transport of structure. Lets stick with your example for now. Suppose, we know that $(M_1,+,\times)$ is a ring and we have a set $M_2$ equipped with two operations $\dot+$ and $\dot\times$. Without further investigation we do not know anything else about $(M_2,\dot+,\dot\times)$ rather than being a set equipped with two operations.
Now, suppose we can define a function $f:M_1\to M_2$ which satisfies the following properties


*

*$f$ is a bijection of sets, saying that $M_1$ and $M_2$ are basically the same set up to renaming

*$f$ satisfies the axioms of a ring homomorphism, i.e. we have $\forall x,y\in M_1$ that
\begin{align*}
f(x+y)&=f(x)\dot+f(y)\\
f(x\times y)&=f(x)\dot\times f(y)
\end{align*}
Then we can transport the given ring structure from $M_1$ to $M_2$ via the function $f:M_1\to M_2$. For example, take the (left) distributive law in $M_2$. Via the surjectivity of $f$ and the ring structure of $M_1$ we may conclude
\begin{align*}
\dot x\dot\times(\dot y\dot+\dot z)&=f(x)\dot\times(f(y)\dot+f(z))\\
&=f(x)\dot\times f(y+z)\\
&=f(x\times (y+z))\\
&=f(x\times y+x\times z)\\
&=f(x\times y)\dot+f(y\times z)\\
&=f(x)\dot\times f(y)\dot+f(y)\dot\times f(z)\\
&=\dot x\dot\times\dot y\dot+\dot x\dot\times\dot z
\end{align*}
In a similiar manner you can in fact prove that all the ring axioms hold in $M_2$ simply by using the properties of the given map $f$ over and over again. I just took the (left) distributive law as an illustration. The same idea applies when $(M_1,+,\times)$ is a field, or a different algebraic structure defined as a set equipped with operations.

But $f$ is not an isomorphism in the first place! The concept of an isomorphism does not apply in this setting as an isomorphism is commonly defined between objects of the same category (as Maryam Ajorlou pointed out). In this setting we use that we have a bijective, structure-preserving map to transport (or define) structure from $M_1$ to $M_2$.

A: What you say is technically correct, but isomorphic doesn't just mean $(M_2,+,\times)$ is a ring, but with the exact same structures. $M_1$ and $M_2$ can be mapped one-to-one on eachother. 
