Below we compare the related forms. First is the iterated descent $\,a\to 103\bmod a\,$ used by Gauss. Second is that rearranged into the form of descending multiples of $60.\,$ Third is the fractional view, and fourth is the graph of the descending multiples of $60$ (denominator descent graph).
$$\begin{align}
103\bmod{60} &= 103 - 1(60) = 43\\
103\bmod 43 &= 103\color{#0a0}{-2(43)=17}\\
103\bmod 17 &= 103-6(17) = 1
\end{align}\qquad\qquad\quad$$
$$\begin{array}{rl}
\bmod{103}\!:\qquad\ (-1)60\!\!\!\! &\equiv\, 43 &\Rightarrow\ 1/60\equiv -1/43\\[.3em]
\smash[t]{\overset{\large\color{#0a0}{*(-2)}}\Longrightarrow}\ \ \ \ \ \ \ \ \ \ (-2)(-1)60\!\!\!\! &\equiv \color{#0a0}{(-2)43\equiv 17}\!\! &\Rightarrow\ 1/60\equiv\ \ \ 2/17\\[.3em]
\smash[t]{\overset{\large *(-6)}\Longrightarrow}\ \ \color{#c00}{(-6)(-2)(-1)}60\!\!\!\! &\equiv (-6)17\equiv 1 &\Rightarrow\ 1/60 \equiv {\color{#c00}{-12}}/1\\
\end{array}$$
$$ \begin{align}
&\dfrac{1}{60}\ \,\equiv\ \ \dfrac{-1}{43}\, \ \equiv\, \ \dfrac{2}{17}\, \equiv\, \dfrac{\color{#c00}{-12}}1\ \ \ \rm[Gauss's\ algorithm]\\[.3em]
&\, 60\overset{\large *(-1)}\longrightarrow\color{#0a0}{43}\overset{\large\color{#0a0}{*(-2)}}\longrightarrow\,\color{#0a0}{17}\overset{\large *(-6)}\longrightarrow 1\\[.4em]
\Rightarrow\ \ &\,60*(-1)\color{#0a0}{*(-2)}*(-6)\equiv 1\ \Rightarrow\ 60^{-1}\rlap{\equiv (-1)(-2)(-6)\equiv \color{#c00}{-12}}
\end{align}$$
The translation from the first form (iterated mods) to the second (iterated smaller multiples) is realized by viewing the modular reductions as modular multiplications, e.g.
$$\ 103\color{#0a0}{-2(43) = 17}\,\Rightarrow\, \color{#0a0}{-2(43) \equiv 17}\!\!\pmod{\!103} $$
This leads to the following simple recursive algorithm for computing inverses $\!\bmod p\,$ prime.
$\begin{align}\rm I(a,p)\ :=\ &\rm if\ \ a = 1\ \ then\ \ 1\qquad\qquad\ \ \ ; \ \ a^{-1}\bmod p,\,\ {\rm for}\ \ a,p\in\Bbb N\,\ \ \&\,\ \ 0 < a < p\ prime \\[.5em]
&\rm else\ let\ [\,q,\,r\,]\, =\, p \div a\qquad ;\, \ \ p = q a + r\ \Rightarrow \color{#0a0}{-qa\,\equiv\, r}\!\!\pmod{\!p},\ \ 0 < r < a\,\\[.2em]
&\rm\ \ \ \ \ \ \ \ \ ({-}q*I(r,p))\bmod p\ \ \ ;\ \ because\ \ \ \dfrac{1}a \equiv \dfrac{-q}{\color{#0a0}{-qa}}\equiv \dfrac{-q}{\color{#0a0}r}\equiv -q * I(r,p)\ \ \ \ \ \color{#90f}{[\![1]\!]} \end{align}
$
Theorem $\ \ {\rm I(a,p)} = a^{-1}\bmod p$
Proof $\ $ Clear if $\,a = 1.\,$ For $\,a > 1,\,$ suppose for induction the theorem holds true for all $\,n < a$. Since $\,p = qa+r\,$ we must have $\,r > 0\,$ (else $\,r = 0\,\Rightarrow\,a\mid p\,$ and $\,1< a < p,\,$ contra $\,p\,$ prime). Thus $\,0 < r < a\,$ so induction $\,\Rightarrow\,{\rm I(r,p)}\equiv \color{#0a0}{r^{-1}}$ so reducing equation $\color{#90f}{[\![1]\!]}\bmod p\,$ yields the claim.
Best Answer
I'm not sure if there is a general consensus, but I would like to add another interpretation different from the ones already mentioned.
I tend to say it is
If not a number nor a congruence class, then what should it be?
I'd say it isn't a thing, it's just a notation used in congruences representing an unspecified inverse of $b$ modulo $m$. I.e, $\frac ab\equiv c\pmod m$ is a notation meaning as much as $a\equiv bc\pmod m$ (provided $\gcd(b,m)=1$), but $\frac ab$ has no meaning to me when pulled out the congruence.
Note that this point of view still conserves the expected properties suggested by the rational numbers, such as $b\cdot\frac ab\equiv a\pmod m$ and $\frac ab+\frac cd\equiv\frac{ad+bc}{bd}\pmod m$ (provided the inverses exist).