Could somebody please help me prove $\frac{a}1=a$ using the properties of real numbers introduced in elementary algebra?

The real numbers form a field; so, lets prove the proposition for an arbitrary field.

Here's the basic facts we know about the multiplicative structure of any field, where $x'$ is notation for the reciprocal of $x$.

1. $\forall xyz : (xy)z=x(yz)$
2. $\forall x(1x = x),\;\; \forall x(x1=x)$
3. $\forall x(x \neq 0 \rightarrow xx' = 1),\;\;\forall x(x \neq 0 \rightarrow x' x = 1)$
4. $\forall xy : xy=yx$
5. $1 \neq 0$

Three comments:

Firstly, you'll note that there is redundancy in the above list (in particular, each version of axiom 2 follows from the other, given 4; similarly, each version of axiom 3 follows from the other, given 4). But that's cool; you don't always have to define things a minimal way. In fact, the pursuit of minimalism can sometimes make things harder than they need to be. That said, you should always search for minimal characterizations after defining something.

Secondly, You may be more familiar with the "existential" version of axioms 2 and 3, namely

$$\exists 1\forall x (x1 = x \wedge 1x=x \wedge \forall x(x \neq 0 \rightarrow \exists y(xy=1)))$$

respectively. I think this is a perfect example of how the pursuit of minimalism can overcomplicate things. Much better the to include reciprocation and the multiplicative identity as part of the data, since this involves less fooling around. Furthermore, (this comment will probably go over your head though) specifying the additional data (in this case, the multiplicative identity $1$ and the reciprocation function) typically gives us a better idea of what homomorphisms of our structures ought to be, especially if this additional data allows us to express our axioms in a weaker fragment of first-order logic (which in this case it does, since there is no existential quantification in the approach I am advocating).

Thirdly, you may be interested to know that, as it turns out, not every structure satisfying the five given axioms can occur as the multiplicative structure of a field. Basically, this is because (but this may also go over your head) there may not exist a suitable Abelian group structure on the underlying set that interacts correctly with multiplication such that the overall result is a field.

Onwards.

Let $F$ denote a field. Your question is (basically): if $a \in F$, how do we know that $a/1 = a$?

First, lets define division. Consider $a,b \in F$ with $b \neq 0$. Then $b'$ is well-defined. Thus by 4, we see that $ab'=b'a$. So lets define that for all $a$ and all non-zero $b$, we have that $$\frac{a}{b}$$ equals either of these two expressions, and therefore both.

Note in particular that since $1 \neq 0$, we have that $\frac{a}{1}$ is always well-defined, for all $a \in F$.

Okay, lets go ahead and prove your result. Let $a \in F$ be fixed but arbitrary. Then to show $\frac{a}{1} = a$, we may argue as follows.

$$\frac{a}{1} = a1' = a1 = a$$

The only step that isn't clear is $1' = 1,$ so lets verify this.

We know from axiom 3 that

$$\forall x(x \neq 0 \rightarrow xx'=1)$$

Thus we may deduce that $$1 \neq 0 \rightarrow 11' = 1.$$

But the hypothesis of the above implication holds, since its just axiom 5. Thus

$$11' = 1.$$

Finally, recall that axiom 2 reads

$$\forall x(1x=x)$$

Thus $$11'=1',$$ and therefore $1 = 1',$ as required.

$\frac{a}{1}=a*\frac{1}{1}$ because $\frac{a}{b}=a*\frac{1}{b}$.

$1*\frac{1}{1}=1$ because $a*\frac{1}{a}=1$.

$1*1=1$ because $a*1=a$.

Comparing the two equations above: $\frac{1}{1}=1$.

$\therefore$ $\frac{a}{1}=a*\frac{1}{1}=a*1=a$.

The set $\mathbb{R}$ of the real numbers is a field $F$ meaning that addition and multiplication are defined and have the usual properties. According to the above link in Wikipedia division is defined implicitly in terms of the inverse operation of multiplication. Among the axioms that are used to give a formal definition of an (additive and multiplicative) field, we will make use of the following two in order to prove the required result:

1. Existance of multiplicative identity element. There is an element, called the multiplicative identity element and denoted by $1$, such that for all $a$ in $F$, $$a \cdot 1 = a$$

2. Existence of multiplicative inverse. For any a in $F$ other than $0$, there exists an element $a^{−1}$ in $F$, such that $$a \cdot a^{-1} = 1$$ The elements $a \cdot b^{−1}$ are also denoted by $\frac{a}{b}$. In other words division operation exists.

So, returning to the question, by substituting $a=1$ in the first property we have that $$1\cdot 1=1$$ which implies based on the second property that $$1^{-1}=1$$ i.e. that $1$ is the multiplicative inverse of itself (which if assumed known, then all the preceding can be omitted). Using the notation in bold (second property) we have that $$\frac{a}{1}=a\cdot 1^{-1}=a\cdot1=a$$ which proves the required equality.