I think, one of the most useful concepts about rings is something I would like to call the "unital closures." (I refuse to use the term "rngs" and, in my post, the term "rings" includes both unital rings and nonunital rings.) For any ring $R$, the "unital closure" is the ring $\hat{R}:=\mathbb{Z}\oplus R$ with the addition $(k,r)+(l,s):=(k+l,r+s)$ and the multiplication $(k,r)\cdot(l,s):=(kl,ks+lr+rs)$ for $k,l\in\mathbb{Z}$ and $r,s\in R$. Clearly, $\hat{R}$ is a unital ring with the multiplicative identity $\left(1,0_R\right)$.
For subsets $A,B$ of $R$, write $A\cdot B$ for the set containing all sums of the form $\sum_{i=1}^n \,a_ib_i$, where $n\in\mathbb{N}$, $a_i\in A$, and $b_i\in B$ for all $i$. A left ideal of $R$ is then an additive abelian subgroup $L$ of $R$ such that $R\cdot L\subseteq I$. If $L$ is generated by a subset $X$ of $R$, then $L$ consists of all elements of the form $\sum_{i=1}^p\,k_ix_i+\sum_{j=1}^q\,r_jy_j$, where $p,q\in\mathbb{N}$, $k_i\in\mathbb{Z}$, $r_j\in R$, and $x_i,y_j\in X$ for every $i,j$. Hence, unlike in the case of unital rings, it is possible that $L\neq R\cdot X$. A two-sided ideal of $R$ is a left ideal $I$ of $R$ such that $I\cdot R \subseteq I$. If $I$ is generated by a subset $X$ of $R$, then $I$ consists of all elements of the form $\sum_{i=1}^p\,k_ix_i+\sum_{j=1}^q\,r_jy_j+\sum_{k=1}^m\,z_ks_k+\sum_{l=1}^n\,t_lw_lu_l$, where $p,q,m,n\in\mathbb{N}$, $k_i\in\mathbb{Z}$, $r_j,s_k,t_l,u_l\in R$, and $x_i,y_j,z_k,w_l\in X$ for every $i,j,k,l$. Again, it is possible that $I\neq R\cdot X\cdot R$, unless $R$ is unital.
A ring homomorphism $\phi:R\to S$ for rings $R$ and $S$ is defined to only be additive and multiplicative. If $R$ and $S$ are unital, then $\phi$ is "unitary" if $\phi\left(1_R\right)=1_S$. Isomorphism theorems for unitary ring homomorphisms carry over to isomorphisms theorems of homomorphisms of potentially nonunital rings. Furthermore, $R$ can be viewed as a subring of $\hat{R}$ via the embedding $r\mapsto (0,r)$ for all $r\in R$. Anyway, any ring homomorphism $\phi:R\to S$ can be extended uniquely to a unitary ring homomorphism $\hat{\phi}:\hat{R}\to\hat{S}$ via $\hat{\phi}\big((k,r)\big):=\big(k,\phi(r )\big)$ for every $k\in\mathbb{Z}$ and $r\in R$. Hence, in a sense, $\mathrm{Hom}(R,S)=\mathrm{Hom}_{\text{unitary}}\left(\hat{R},\hat{S}\right)$. A good thing about taking nonunital rings into account is that the category $\mathbf{Rings}$ of all rings together with ring homomorphisms has a zero object (the zero ring), but the category $\mathbf{URings}$ of all unital rings together with unitary ring homomorphisms has no zero objects, but an initial object ($\mathbb{Z}$) and a terminal object (again, the zero ring).
A left $R$-module $M$ is defined in the usual way, and if $R$ is unital, we say that $M$ is "unitary" if $1_R\cdot m=m$ for all $m\in M$. Homomorphisms of left $R$-modules are also as usual. I shall denote by $\mathbf{LMod}(R )$ the category of left $R$-modules together with homomorphisms of left $R$-modules, and by $\mathbf{LUMod}(U )$ the category of unitary left $U$-modules together with homomorphisms of left $U$-modules, given that $U$ is a unital ring. Obviously, $\mathbf{LUMod}(U )$ is a full subcategory of $\mathbf{LMod}(U)$ for every unital ring $U$.
Another essential thing about the unital closure $\hat{R}$ of a ring $R$ is that $\mathbf{LMod}(R )$ is identical to $\mathbf{LUMod}\left(\hat{R}\right)$ (since every left $R$-module $M$ is a left $\hat{R}$-module via $(k,r)\cdot m :=km+r\cdot m$ for $k\in\mathbb{Z}$, $r\in R$, and $m\in M$, and every left $\hat{R}$-module $N$ is a left $R$-module via $r\cdot n:=(0,r)\cdot n$ for all $r\in R$ and $n\in N$). Hence, many properties of unitary modules over a unital ring can be generalized to modules over a ring. For example, you know that a unitary free module over a unitary ring $U$ is a direct sum of copies of $U$. What about a free module over $R$? Well, as $\hat{R}$ is unital, a free module over $R$ then is a unitary free module over $\hat{R}$, hence it will be a direct sum of copies of $\hat{R}=\mathbb{Z}\oplus R$. Another example is that $\mathbf{LMod}(R )$ has enough projectives and enough injectives, which is due to the fact that $\mathbf{LUMod}(U)$ has enough projectives and enough injectives.
Now, I return again to Clément Guérin's answer about ideals of rings. For a ring $R$, a left ideal of $R$ is actually a left $\hat{R}$-submodule of $R$, where $R$ is given the left $\hat{R}$-module structure $(k,r)\cdot s:=ks+rs$ for $k\in\mathbb{Z}$ and $r,s\in R$. So, the left ideal $\langle a \rangle$ generated by $a\in R$ will be $\left\{ka+ra\,|\,k\in\mathbb{Z}\text{ and }r\in R\right\}$, which is equal to $R\cdot a$ if $R$ is unital.
PS: I omitted the discussion of right ideals and right modules for a trivial reason.
I just remembered a problem I dealt with recently.
Let $R$ be a ring and $n\in\mathbb{N}$. What are all left ideals and two-sided ideals of the ring of matrices $S:=\mathrm{Mat}_{n\times n}(R)$?
Assume that $R$ is unital. Then, there is a one-to-one correspondence between the set $\mathcal{L}\left(R^n\right)$ of all left $R$-submodules of the unitary $R$-module $R^n$ and the set $\mathcal{L}(S)$ of all left ideals of $S$ which associates each $V\in\mathcal{L}\left(R^n\right)$ with the left ideal of $S$ consisting of matrices $\left[v_{i,j}\right]_{i,j\in[n]}$, where $[n]:=\{1,2,\ldots,n\}$, and $\left(v_{i,1},v_{i,2},\ldots,v_{i,n}\right) \in V$ for every $i\in[n]$. There is also a one-to-one correspondence between the set $\mathcal{T}(R)$ of two-sided ideal of $R$ and the set $\mathcal{T}(S)$ of two-sided ideals of $S$. This correspondence associates $I\in\mathcal{T}(R)$ with $\text{Mat}_{n\times n}(I)$.
Now, what happens if $R$ is nonunital? I don't know the answer, and I suspect that it is an open question. I don't even know the answer when $R$ is a trivial ring (i.e., an additive abelian group $R$ with the trivial multiplication: $r\cdot s:=0_R$ for all $r,s\in R$).
